For each scenario, name the quantity you neither know nor need, say which equation that points to, and then find the requested value. 1. A rocket sled starts from rest and accelerates uniformly at 8.0 m/s^2. How fast is it moving after it has traveled 240 m? 2. A stone is thrown straight down from a bridge at 5.0 m/s. How far has it fallen 1.5 s later? 3. A train slows uniformly from 30.0 m/s to 18.0 m/s in 8.0 s. How far does it travel in that time?

1. Missing t, so use the no-time equation: v_x = 62.0 m/s. 2. Missing v_x, so use the position equation: 18.5 m. 3. Missing a_x, so use the average-velocity equation: 192 m.

A skateboarder is rolling down a straight ramp at 3.0 m/s when she tucks and picks up a constant acceleration of 1.4 m/s^2 in her direction of travel. What is her velocity 4.5 s later?

v_x = 9.3 m/s, still directed down the ramp.

A cyclist coasting at 12.0 m/s applies the brakes and slows uniformly to 4.0 m/s with an acceleration of magnitude 1.6 m/s^2 opposite his motion. How long does the braking take, and how far does he travel while braking?

Braking takes 5.0 s and covers 40.0 m.

A car merging onto a highway speeds up uniformly from 11.0 m/s to 27.0 m/s while covering 152 m of the on-ramp. Find its acceleration.

a_x = 2.00 m/s^2, in the direction of travel.

A wrench slips off a scaffold and falls freely from rest, 14.5 m above the ground. How long is it in the air, and how fast is it moving just before it lands?

The wrench falls for 1.72 s and lands at 16.9 m/s, moving downward.

A ball is thrown straight up at 21.0 m/s from the point where it leaves the thrower's hand. Find (a) the time to reach the highest point, (b) the maximum height above the release point, (c) the total time until it falls back to the release point, and (d) its velocity at that moment.

(a) 2.14 s to the top. (b) 22.5 m above the release point. (c) 4.29 s for the round trip, exactly double the time up. (d) -21.0 m/s, that is 21.0 m/s downward.

A cart moves along a straight track. Its velocity-time graph is three straight segments: velocity rises steadily from 2.0 m/s at t = 0 to 11.0 m/s at t = 3.0 s, holds at 11.0 m/s until t = 7.0 s, then falls steadily to 1.0 m/s at t = 12.0 s. Velocity is positive throughout. Find the acceleration on each segment, the total displacement, and the average velocity for the 12.0 s.

Accelerations 3.0 m/s^2, then 0, then -2.0 m/s^2. Total displacement 93.5 m, average velocity 7.79 m/s.

A commuter train leaves a station from rest and accelerates uniformly at 1.2 m/s^2 until it reaches 24.0 m/s. It then holds that speed for a further 90.0 s. Find the total distance covered and the average speed for the whole trip.

Total distance 2400 m (2.40 km) in 110.0 s, an average speed of 21.8 m/s.

A driver travelling at 22.0 m/s sees a fallen branch in the road. She holds a constant speed for a reaction time of 0.65 s, then brakes with a constant acceleration of magnitude 5.5 m/s^2 until the car stops. The branch is 60.0 m ahead of the car at the instant she sees it. Does she stop in time, and by how much?

Total stopping distance 58.3 m (14.3 m of it before the brakes act), so she stops about 1.7 m short of the branch.

A student argues: a ball dropped from rest falls 4.9 m during the first second, and gravity adds 9.8 m/s of speed every second, so the ball must fall 9.8 m during the second second. Is the student right? Work out the distance fallen during each of the first three seconds.

No. The distances are 4.9 m, 14.7 m and 24.5 m, a 1 : 3 : 5 pattern. The 9.8 m/s per second figure is the rate the velocity grows, not the distance.

A classmate says there are four kinematic equations. You have counted three on the AP Physics 1 equation sheet. Who is right, where does the fourth come from, and what condition does it need? Illustrate with a runner who speeds up uniformly from 2.0 m/s to 8.0 m/s in 4.0 s.

Both. The sheet prints three; the standard textbook set is four. The fourth, delta x = (v_0 + v)/(2) t, follows from the printed position equation plus a t = v - v_0, and holds only for constant acceleration. For the runner, delta x = 20.0 m by either route.

12 Kinematics Practice Problems With Full Solutions

Q: What are the four kinematic equations?

The four standard constant-acceleration equations are v = v_0 + a t, \Delta x = v_0 t + \frac{1}{2} a t^2, v^2 = v_0^2 + 2 a \Delta x, and \Delta x = \frac{v_0 + v}{2} t. The AP Physics 1 equation sheet prints the first three and leaves out the fourth, so on exam day you either recall it or derive it by noting that displacement equals average velocity times time and that, for constant acceleration, average velocity is the plain mean of the starting and ending velocities. All four require the acceleration to be constant over the interval you apply them to.

Twelve constant-acceleration problems, easiest first, each with the whole solution shown. The skill they all train is picking the equation that leaves out the one variable you neither know nor need. Every problem states its positive direction before the first number and uses g = 9.8.

AP Physics: Unit 1 (topics 1.2 Displacement, Velocity, and Acceleration, 1.3 Representing Motion). Composed practice for AP Physics 1 Unit 1, Topics 1.2 and 1.3. The three constant-acceleration equations used throughout are the ones listed in essential knowledge 1.3.A.2 and printed on the equation sheet; the graph problem uses the slope and area readings in 1.3.A.4.ii and 1.3.A.4.iii. These are original problems, not released College Board items.

What this set covers

Twelve problems on motion in a straight line with constant acceleration, ordered from a single plug-in to a two-phase trip. Between them they cover the shapes that actually turn up:

Choosing an equation from the variable that is missing.
Solving for final velocity, for time, for displacement, and for acceleration.
Free fall: a dropped object, and an object thrown straight up.
Reading a velocity-time graph, where the slope is acceleration and the area is displacement.
A two-phase trip and a reaction-time-plus-braking problem, where you solve one phase at a time and add.
Two questions whose answer is reasoning rather than a number.

Motion in two dimensions gets its own set: see the projectile motion practice problems, which apply these same equations one axis at a time. For the method behind them, read the kinematic equations guide first; this page is the drilling, not the teaching. The CED framing for the definitions is on Topic 1.2, and the rest of the unit is indexed on the Unit 1 overview.

The equations, and which ones the sheet prints

The AP Physics 1 equation sheet prints exactly three constant-acceleration equations for linear motion:

v_x = v_{x0} + a_x t

x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2

v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)

The CED says the same thing in words. Essential knowledge 1.3.A.2 reads that for constant acceleration, three kinematic equations can be used to describe instantaneous linear motion in one dimension, and prints those three, with the note that the equations are written for the x-direction but can be used in any single dimension as appropriate.

A fourth is standard in textbooks and is not on the sheet:

\Delta x = \frac{v_0 + v}{2} \, t

It is displacement equals average velocity times time, using the fact that average velocity equals the plain mean of the endpoints only when acceleration is constant. You may use it, but you have to supply it yourself. Problem 12 derives it in two lines.

Five quantities appear across the four equations: displacement, initial velocity, final velocity, acceleration, and time. Each equation omits exactly one of them, which is what makes the choice mechanical.

The variable you neither know nor need	Use	On the AP sheet?
Displacement $(x - x_0)$	$v_x = v_{x0} + a_x t$	Yes
Final velocity $(v_x)$	$x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2$	Yes
Time $(t)$	$v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)$	Yes
Acceleration $(a_x)$	$\Delta x = \frac{v_0 + v}{2} t$	No

The sheet also prints the three rotational analogues, with $\omega$ , $\theta$ and $\alpha$ in place of $v$ , $x$ and $a$ . Those belong to Unit 5 and are not used here.

Conventions, and how to work through the set

Every problem below follows the same rules, so you can compare your work line for line:

$g = 9.8 \ \mathrm{m/s^2}$ , the value printed in the AP Physics 1 Table of Information. The CED also states $g \approx 10 \ \mathrm{m/s^2}$ and says the exam uses that value where a number is required, while adding that you are not penalized for correctly using 9.81 or 9.8. Problem answers here would shift in the second digit with $g = 10$ .
The positive direction is stated before the first number and held to the last line. If a solution below flips an axis midway, that is a bug, not a shortcut.
Air resistance is negligible, which is the standing convention listed on the AP Physics 1 equation sheet unless a problem says otherwise.
Intermediate values are carried to one more digit than the answer, then rounded once at the end.

Attempt each problem with the solution closed. If your number differs, open the solution and find the first line you disagree with rather than reading it straight through. Then put the same inputs into the kinematics calculator, which shows the equation it used, and check the graph reading against Topic 1.3.

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. Choose the equation before you calculate
For each scenario, name the quantity you neither know nor need, say which equation that points to, and then find the requested value.
1. A rocket sled starts from rest and accelerates uniformly at $8.0 \ \mathrm{m/s^2}$ . How fast is it moving after it has traveled 240 m?
2. A stone is thrown straight down from a bridge at $5.0 \ \mathrm{m/s}$ . How far has it fallen 1.5 s later?
3. A train slows uniformly from $30.0 \ \mathrm{m/s}$ to $18.0 \ \mathrm{m/s}$ in 8.0 s. How far does it travel in that time?
Show the worked solution
1. Scenario 1. Take the direction of travel as positive. Knowns: $v_{x0} = 0$ , $a_x = 8.0 \ \mathrm{m/s^2}$ , $x - x_0 = 240 \ \mathrm{m}$ . Time is never mentioned and never asked for, so the missing variable is $t$ and the equation is $v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)$ .
2. Substitute: $v_x^2 = 0 + 2(8.0 \ \mathrm{m/s^2})(240 \ \mathrm{m}) = 3840 \ \mathrm{m^2/s^2}$ , so $v_x = \sqrt{3840} = 62.0 \ \mathrm{m/s}$ .
3. Scenario 2. Take down as positive, so gravity is $+9.8 \ \mathrm{m/s^2}$ and the throw is $+5.0 \ \mathrm{m/s}$ . Knowns: $v_{x0} = 5.0 \ \mathrm{m/s}$ , $a_x = 9.8 \ \mathrm{m/s^2}$ , $t = 1.5 \ \mathrm{s}$ . The speed at the end is neither given nor wanted, so the missing variable is $v_x$ and the equation is $x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2$ .
4. Substitute: $x - x_0 = (5.0)(1.5) + \frac{1}{2}(9.8)(1.5)^2 = 7.5 \ \mathrm{m} + 11.025 \ \mathrm{m} = 18.525 \ \mathrm{m}$ , which is $18.5 \ \mathrm{m}$ to three figures.
5. Scenario 3. Take the direction of travel as positive. Knowns: $v_0 = 30.0 \ \mathrm{m/s}$ , $v = 18.0 \ \mathrm{m/s}$ , $t = 8.0 \ \mathrm{s}$ . Acceleration is neither given nor asked for, so the missing variable is $a_x$ , and the only equation without it is the one the sheet does not print: $\Delta x = \frac{v_0 + v}{2} t$ .
6. Substitute: $\Delta x = \frac{30.0 + 18.0}{2}(8.0) = (24.0 \ \mathrm{m/s})(8.0 \ \mathrm{s}) = 192 \ \mathrm{m}$ .
7. Sheet-only route for scenario 3, if you would rather not rely on the fourth equation: $a_x = \frac{18.0 - 30.0}{8.0} = -1.5 \ \mathrm{m/s^2}$ , then $\Delta x = (30.0)(8.0) + \frac{1}{2}(-1.5)(8.0)^2 = 240 \ \mathrm{m} - 48 \ \mathrm{m} = 192 \ \mathrm{m}$ . Same answer, one extra step.
1. Missing $t$ , so use the no-time equation: $v_x = 62.0 \ \mathrm{m/s}$ . 2. Missing $v_x$ , so use the position equation: $18.5 \ \mathrm{m}$ . 3. Missing $a_x$ , so use the average-velocity equation: $192 \ \mathrm{m}$ .
2. Final velocity from an acceleration and a time
A skateboarder is rolling down a straight ramp at $3.0 \ \mathrm{m/s}$ when she tucks and picks up a constant acceleration of $1.4 \ \mathrm{m/s^2}$ in her direction of travel. What is her velocity 4.5 s later?
Show the worked solution
1. Take the direction of travel down the ramp as positive. Knowns: $v_{x0} = 3.0 \ \mathrm{m/s}$ , $a_x = 1.4 \ \mathrm{m/s^2}$ , $t = 4.5 \ \mathrm{s}$ . Target: $v_x$ .
2. Displacement is neither given nor asked for, so use the equation missing $(x - x_0)$ : $v_x = v_{x0} + a_x t$ .
3. Substitute: $v_x = 3.0 \ \mathrm{m/s} + (1.4 \ \mathrm{m/s^2})(4.5 \ \mathrm{s}) = 3.0 \ \mathrm{m/s} + 6.3 \ \mathrm{m/s} = 9.3 \ \mathrm{m/s}$ .
4. Sense check: gaining $1.4 \ \mathrm{m/s}$ every second for four and a half seconds is a gain of a little over $6 \ \mathrm{m/s}$ , and the sign is positive, so she is still heading down the ramp.
$v_x = 9.3 \ \mathrm{m/s}$ , still directed down the ramp.
3. Displacement from rest
A subway train pulls out of a station from rest and accelerates uniformly at $0.75 \ \mathrm{m/s^2}$ for 12.0 s. How far does it travel in that time?
Show the worked solution
1. Take the direction of travel as positive. Knowns: $v_{x0} = 0$ (from rest), $a_x = 0.75 \ \mathrm{m/s^2}$ , $t = 12.0 \ \mathrm{s}$ . Target: $x - x_0$ .
2. The velocity at the end of the 12.0 s is neither given nor asked for, so use the equation missing $v_x$ : $x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2$ .
3. With $v_{x0} = 0$ the first term drops out: $x - x_0 = \frac{1}{2}(0.75 \ \mathrm{m/s^2})(12.0 \ \mathrm{s})^2 = \frac{1}{2}(0.75)(144) = 54.0 \ \mathrm{m}$ .
4. Sense check with average velocity: the train ends at $v_x = (0.75)(12.0) = 9.0 \ \mathrm{m/s}$ , so its average speed over the interval is $4.5 \ \mathrm{m/s}$ and $(4.5 \ \mathrm{m/s})(12.0 \ \mathrm{s}) = 54.0 \ \mathrm{m}$ . Agreed.
The train travels $54.0 \ \mathrm{m}$ .
4. Time for a cyclist to slow down
A cyclist coasting at $12.0 \ \mathrm{m/s}$ applies the brakes and slows uniformly to $4.0 \ \mathrm{m/s}$ with an acceleration of magnitude $1.6 \ \mathrm{m/s^2}$ opposite his motion. How long does the braking take, and how far does he travel while braking?
Show the worked solution
1. Take the direction of motion as positive, which makes the braking acceleration negative. Knowns: $v_{x0} = 12.0 \ \mathrm{m/s}$ , $v_x = 4.0 \ \mathrm{m/s}$ , $a_x = -1.6 \ \mathrm{m/s^2}$ . First target: $t$ .
2. Displacement is not yet needed, so use the equation missing it: $v_x = v_{x0} + a_x t$ , rearranged to $t = \frac{v_x - v_{x0}}{a_x}$ .
3. Substitute: $t = \frac{4.0 - 12.0}{-1.6} = \frac{-8.0 \ \mathrm{m/s}}{-1.6 \ \mathrm{m/s^2}} = 5.0 \ \mathrm{s}$ . Two negatives give a positive time, which is the sign check that the setup is consistent.
4. Second target: distance. Now that $t$ is known, $\Delta x = v_{x0} t + \frac{1}{2} a_x t^2 = (12.0)(5.0) + \frac{1}{2}(-1.6)(5.0)^2 = 60.0 \ \mathrm{m} - 20.0 \ \mathrm{m} = 40.0 \ \mathrm{m}$ .
5. Check the distance a second way, with the average-velocity equation: $\Delta x = \frac{12.0 + 4.0}{2}(5.0) = (8.0)(5.0) = 40.0 \ \mathrm{m}$ . Agreed.
Braking takes $5.0 \ \mathrm{s}$ and covers $40.0 \ \mathrm{m}$ .
5. Acceleration from two speeds and a distance
A car merging onto a highway speeds up uniformly from $11.0 \ \mathrm{m/s}$ to $27.0 \ \mathrm{m/s}$ while covering 152 m of the on-ramp. Find its acceleration.
Show the worked solution
1. Take the direction of travel as positive. Knowns: $v_{x0} = 11.0 \ \mathrm{m/s}$ , $v_x = 27.0 \ \mathrm{m/s}$ , $x - x_0 = 152 \ \mathrm{m}$ . Target: $a_x$ .
2. The time on the ramp is neither given nor asked for, so use the equation missing $t$ : $v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)$ , rearranged to $a_x = \frac{v_x^2 - v_{x0}^2}{2(x - x_0)}$ .
3. Substitute: $a_x = \frac{(27.0)^2 - (11.0)^2}{2(152)} = \frac{729 - 121}{304} = \frac{608 \ \mathrm{m^2/s^2}}{304 \ \mathrm{m}} = 2.00 \ \mathrm{m/s^2}$ .
4. Check by finding the time you did not need: $t = \frac{27.0 - 11.0}{2.00} = 8.0 \ \mathrm{s}$ , and then $\Delta x = (11.0)(8.0) + \frac{1}{2}(2.00)(8.0)^2 = 88 \ \mathrm{m} + 64 \ \mathrm{m} = 152 \ \mathrm{m}$ . Back to the given distance.
$a_x = 2.00 \ \mathrm{m/s^2}$ , in the direction of travel.
6. A dropped wrench
A wrench slips off a scaffold and falls freely from rest, 14.5 m above the ground. How long is it in the air, and how fast is it moving just before it lands?
Show the worked solution
1. Take down as positive, so every quantity in this problem is positive and no minus signs appear. Knowns: $v_{x0} = 0$ (it slips rather than being thrown), $a_x = 9.8 \ \mathrm{m/s^2}$ , $x - x_0 = 14.5 \ \mathrm{m}$ . First target: $t$ .
2. The landing speed is not yet known, so use the equation missing $v_x$ : $x - x_0 = v_{x0} t + \frac{1}{2} a_x t^2$ . With $v_{x0} = 0$ this is $14.5 = \frac{1}{2}(9.8) t^2 = 4.9 t^2$ .
3. Solve: $t^2 = \frac{14.5}{4.9} = 2.9592 \ \mathrm{s^2}$ , so $t = 1.7202 \ \mathrm{s}$ , or $1.72 \ \mathrm{s}$ to three figures.
4. Second target: landing speed. Fastest route now that $t$ is known: $v_x = v_{x0} + a_x t = 0 + (9.8)(1.7202) = 16.858 \ \mathrm{m/s}$ , or $16.9 \ \mathrm{m/s}$ .
5. Independent check with the no-time equation: $v_x^2 = 0 + 2(9.8)(14.5) = 284.2 \ \mathrm{m^2/s^2}$ , so $v_x = \sqrt{284.2} = 16.858 \ \mathrm{m/s}$ . The two routes agree to every digit shown, which is the point of doing both.
The wrench falls for $1.72 \ \mathrm{s}$ and lands at $16.9 \ \mathrm{m/s}$ , moving downward.
7. Thrown straight up: signs and symmetry
A ball is thrown straight up at $21.0 \ \mathrm{m/s}$ from the point where it leaves the thrower's hand. Find (a) the time to reach the highest point, (b) the maximum height above the release point, (c) the total time until it falls back to the release point, and (d) its velocity at that moment.
Show the worked solution
1. Take up as positive. That one choice fixes every sign: $v_{x0} = +21.0 \ \mathrm{m/s}$ and $a_x = -9.8 \ \mathrm{m/s^2}$ for the entire flight, going up, at the top, and coming down. Gravity does not switch off or change sign at the peak.
2. (a) At the highest point the velocity is momentarily zero, so $v_x = 0$ . Displacement is not needed yet, so use $v_x = v_{x0} + a_x t$ : $0 = 21.0 - 9.8 t$ , giving $t = \frac{21.0}{9.8} = 2.1429 \ \mathrm{s}$ , or $2.14 \ \mathrm{s}$ .
3. (b) For the height, time is now the variable you can skip, so use $v_x^2 = v_{x0}^2 + 2 a_x (x - x_0)$ with $v_x = 0$ : $0 = (21.0)^2 + 2(-9.8)(x - x_0)$ , so $x - x_0 = \frac{441}{19.6} = 22.5 \ \mathrm{m}$ .
4. (c) The return to the release point means $x - x_0 = 0$ . Substituting into $x - x_0 = v_{x0} t + \frac{1}{2} a_x t^2$ gives $0 = 21.0 t - 4.9 t^2 = t(21.0 - 4.9 t)$ . The root $t = 0$ is the launch itself; the one you want is $t = \frac{21.0}{4.9} = 4.2857 \ \mathrm{s}$ , or $4.29 \ \mathrm{s}$ .
5. Notice that $4.2857 \ \mathrm{s}$ is exactly twice the $2.1429 \ \mathrm{s}$ from part (a). The rise and the fall take the same time, because the trip down is the trip up run backwards. Once you trust that, part (c) is one multiplication.
6. (d) Velocity on return: $v_x = 21.0 + (-9.8)(4.2857) = 21.0 - 42.0 = -21.0 \ \mathrm{m/s}$ . The magnitude matches the launch speed and the minus sign says downward, which is the same symmetry seen from the velocity side.
(a) $2.14 \ \mathrm{s}$ to the top. (b) $22.5 \ \mathrm{m}$ above the release point. (c) $4.29 \ \mathrm{s}$ for the round trip, exactly double the time up. (d) $-21.0 \ \mathrm{m/s}$ , that is $21.0 \ \mathrm{m/s}$ downward.
8. Reading a velocity-time graph
A cart moves along a straight track. Its velocity-time graph is three straight segments: velocity rises steadily from $2.0 \ \mathrm{m/s}$ at $t = 0$ to $11.0 \ \mathrm{m/s}$ at $t = 3.0 \ \mathrm{s}$ , holds at $11.0 \ \mathrm{m/s}$ until $t = 7.0 \ \mathrm{s}$ , then falls steadily to $1.0 \ \mathrm{m/s}$ at $t = 12.0 \ \mathrm{s}$ . Velocity is positive throughout. Find the acceleration on each segment, the total displacement, and the average velocity for the 12.0 s.
Show the worked solution
1. Take the direction of motion as positive; the problem says velocity stays positive, so the cart never reverses. On a velocity-time graph the slope is the acceleration and the area between the line and the time axis is the displacement. Those two readings are essential knowledge 1.3.A.4.ii and 1.3.A.4.iii in the CED.
2. Segment 1, $t = 0$ to $3.0 \ \mathrm{s}$ : $a = \frac{11.0 - 2.0}{3.0 - 0} = \frac{9.0}{3.0} = 3.0 \ \mathrm{m/s^2}$ .
3. Segment 2, $t = 3.0$ to $7.0 \ \mathrm{s}$ : the line is flat, so $a = 0$ . Constant velocity, not zero velocity.
4. Segment 3, $t = 7.0$ to $12.0 \ \mathrm{s}$ : $a = \frac{1.0 - 11.0}{12.0 - 7.0} = \frac{-10.0}{5.0} = -2.0 \ \mathrm{m/s^2}$ . Negative acceleration with positive velocity means slowing down, and the cart is still moving forward at the end.
5. Areas. Segment 1 is a trapezoid: $\frac{2.0 + 11.0}{2}(3.0) = (6.5)(3.0) = 19.5 \ \mathrm{m}$ . Segment 2 is a rectangle: $(11.0)(4.0) = 44.0 \ \mathrm{m}$ . Segment 3 is a trapezoid: $\frac{11.0 + 1.0}{2}(5.0) = (6.0)(5.0) = 30.0 \ \mathrm{m}$ .
6. Total displacement: $19.5 + 44.0 + 30.0 = 93.5 \ \mathrm{m}$ . Every area is above the axis, so distance and displacement are the same here; a segment below the axis would count negative.
7. Average velocity is total displacement over total time, not the average of the three segment speeds: $v_{\text{avg}} = \frac{93.5 \ \mathrm{m}}{12.0 \ \mathrm{s}} = 7.79 \ \mathrm{m/s}$ .
Accelerations $3.0 \ \mathrm{m/s^2}$ , then $0$ , then $-2.0 \ \mathrm{m/s^2}$ . Total displacement $93.5 \ \mathrm{m}$ , average velocity $7.79 \ \mathrm{m/s}$ .
9. Two-phase trip: accelerate, then cruise
A commuter train leaves a station from rest and accelerates uniformly at $1.2 \ \mathrm{m/s^2}$ until it reaches $24.0 \ \mathrm{m/s}$ . It then holds that speed for a further 90.0 s. Find the total distance covered and the average speed for the whole trip.
Show the worked solution
1. Take the direction of travel as positive. This is two problems glued together: the kinematic equations apply to phase 1, where acceleration is constant and nonzero, and to phase 2, where it is constant at zero. Never apply one equation across both.
2. Phase 1, duration. Knowns: $v_{x0} = 0$ , $v_x = 24.0 \ \mathrm{m/s}$ , $a_x = 1.2 \ \mathrm{m/s^2}$ . From $v_x = v_{x0} + a_x t_1$ : $t_1 = \frac{24.0}{1.2} = 20.0 \ \mathrm{s}$ .
3. Phase 1, distance. $\Delta x_1 = \frac{1}{2} a_x t_1^2 = \frac{1}{2}(1.2)(20.0)^2 = \frac{1}{2}(1.2)(400) = 240 \ \mathrm{m}$ . Cross-check with average velocity: $\frac{0 + 24.0}{2}(20.0) = 240 \ \mathrm{m}$ .
4. Phase 2, distance. Acceleration is zero, so displacement is just speed times time: $\Delta x_2 = (24.0 \ \mathrm{m/s})(90.0 \ \mathrm{s}) = 2160 \ \mathrm{m}$ .
5. Totals. Distance $= 240 + 2160 = 2400 \ \mathrm{m}$ , that is $2.40 \ \mathrm{km}$ . Total time $= 20.0 + 90.0 = 110.0 \ \mathrm{s}$ .
6. Average speed $= \frac{2400 \ \mathrm{m}}{110.0 \ \mathrm{s}} = 21.8 \ \mathrm{m/s}$ . It sits below the cruising speed of $24.0 \ \mathrm{m/s}$ , as it must, because the train spent 20 s of the trip slower than that.
Total distance $2400 \ \mathrm{m}$ ( $2.40 \ \mathrm{km}$ ) in $110.0 \ \mathrm{s}$ , an average speed of $21.8 \ \mathrm{m/s}$ .
10. Reaction time plus braking
A driver travelling at $22.0 \ \mathrm{m/s}$ sees a fallen branch in the road. She holds a constant speed for a reaction time of 0.65 s, then brakes with a constant acceleration of magnitude $5.5 \ \mathrm{m/s^2}$ until the car stops. The branch is 60.0 m ahead of the car at the instant she sees it. Does she stop in time, and by how much?
Show the worked solution
1. Take the direction of travel as positive. Split the motion at the moment the brakes bite: constant velocity first, constant negative acceleration second.
2. Reaction phase. Acceleration is zero, so $\Delta x_1 = v t_{\text{react}} = (22.0 \ \mathrm{m/s})(0.65 \ \mathrm{s}) = 14.3 \ \mathrm{m}$ . The car covers this distance before the brakes do anything.
3. Braking phase. Knowns at the start of braking: $v_{x0} = 22.0 \ \mathrm{m/s}$ , $v_x = 0$ , $a_x = -5.5 \ \mathrm{m/s^2}$ . Time is neither given nor asked for, so use $v_x^2 = v_{x0}^2 + 2 a_x \Delta x_2$ .
4. Substitute: $0 = (22.0)^2 + 2(-5.5)\Delta x_2$ , so $\Delta x_2 = \frac{484 \ \mathrm{m^2/s^2}}{11.0 \ \mathrm{m/s^2}} = 44.0 \ \mathrm{m}$ .
5. Total. $14.3 + 44.0 = 58.3 \ \mathrm{m}$ , against 60.0 m available. She stops with $60.0 - 58.3 = 1.7 \ \mathrm{m}$ to spare.
6. Worth noticing how the two phases scale differently. Reaction distance is proportional to speed, braking distance to speed squared. At $22.0 \ \mathrm{m/s}$ the braking part is already three times the reaction part; at twice the speed it would be six times.
7. Time check, if you want it: braking lasts $t_2 = \frac{22.0}{5.5} = 4.0 \ \mathrm{s}$ , so the whole event takes $0.65 + 4.0 = 4.65 \ \mathrm{s}$ .
Total stopping distance $58.3 \ \mathrm{m}$ (14.3 m of it before the brakes act), so she stops about $1.7 \ \mathrm{m}$ short of the branch.
11. Does a falling object cover 9.8 m in the second second?
A student argues: a ball dropped from rest falls 4.9 m during the first second, and gravity adds $9.8 \ \mathrm{m/s}$ of speed every second, so the ball must fall 9.8 m during the second second. Is the student right? Work out the distance fallen during each of the first three seconds.
Show the worked solution
1. Take down as positive, $v_{x0} = 0$ , $a_x = 9.8 \ \mathrm{m/s^2}$ . Distance fallen from release to time $t$ is $d(t) = \frac{1}{2}(9.8) t^2 = 4.9 t^2$ .
2. Cumulative distances: $d(1.0) = 4.9(1.00) = 4.9 \ \mathrm{m}$ , $d(2.0) = 4.9(4.00) = 19.6 \ \mathrm{m}$ , $d(3.0) = 4.9(9.00) = 44.1 \ \mathrm{m}$ .
3. Distance in each individual second is the difference of consecutive cumulative distances. First second: $4.9 - 0 = 4.9 \ \mathrm{m}$ . Second second: $19.6 - 4.9 = 14.7 \ \mathrm{m}$ . Third second: $44.1 - 19.6 = 24.5 \ \mathrm{m}$ .
4. So the student is wrong, and by a lot. The three distances are in the ratio $4.9 : 14.7 : 24.5$ , which is $1 : 3 : 5$ .
5. Where the reasoning went astray: $9.8 \ \mathrm{m/s}$ per second is the gain in velocity, not the gain in distance. During the second second the velocity runs from $9.8 \ \mathrm{m/s}$ to $19.6 \ \mathrm{m/s}$ , so the average velocity over that interval is $14.7 \ \mathrm{m/s}$ , and $(14.7 \ \mathrm{m/s})(1.0 \ \mathrm{s}) = 14.7 \ \mathrm{m}$ . That reproduces the answer above in one line.
6. The same check works for the third second: velocity runs $19.6$ to $29.4 \ \mathrm{m/s}$ , average $24.5 \ \mathrm{m/s}$ , distance $24.5 \ \mathrm{m}$ .
No. The distances are $4.9 \ \mathrm{m}$ , $14.7 \ \mathrm{m}$ and $24.5 \ \mathrm{m}$ , a $1 : 3 : 5$ pattern. The $9.8 \ \mathrm{m/s}$ per second figure is the rate the velocity grows, not the distance.
12. Three equations or four?
A classmate says there are four kinematic equations. You have counted three on the AP Physics 1 equation sheet. Who is right, where does the fourth come from, and what condition does it need? Illustrate with a runner who speeds up uniformly from $2.0 \ \mathrm{m/s}$ to $8.0 \ \mathrm{m/s}$ in 4.0 s.
Show the worked solution
1. Both statements are true about different things. The AP Physics 1 equation sheet prints exactly three constant-acceleration equations for linear motion, and the CED says the same in essential knowledge 1.3.A.2. Textbooks usually teach a set of four. The extra one is real physics; it is simply not supplied to you on exam day.
2. The fourth equation is $\Delta x = \frac{v_0 + v}{2} t$ . Derive it from a sheet equation in two lines. Start with $\Delta x = v_0 t + \frac{1}{2} a t^2$ and write the last term as $\frac{1}{2}(a t) t$ .
3. From the first sheet equation, $a t = v - v_0$ . Substituting gives $\Delta x = v_0 t + \frac{1}{2}(v - v_0) t = \left(v_0 + \frac{v - v_0}{2}\right) t = \frac{v_0 + v}{2} t$ .
4. The condition is constant acceleration, and it enters exactly where you would expect. $\Delta x = v_{\text{avg}} t$ is always true by the definition of average velocity. What needs constant acceleration is the extra claim that $v_{\text{avg}}$ equals the plain mean $\frac{v_0 + v}{2}$ of the two endpoint velocities. If the acceleration changes during the interval, the velocity spends unequal time near each end, and that mean is no longer the average.
5. Runner: $\Delta x = \frac{2.0 + 8.0}{2}(4.0) = (5.0 \ \mathrm{m/s})(4.0 \ \mathrm{s}) = 20.0 \ \mathrm{m}$ .
6. Sheet-only check: $a = \frac{8.0 - 2.0}{4.0} = 1.5 \ \mathrm{m/s^2}$ , then $\Delta x = (2.0)(4.0) + \frac{1}{2}(1.5)(4.0)^2 = 8.0 + 12.0 = 20.0 \ \mathrm{m}$ . Same distance, one step longer.
7. Practical upshot: the fourth equation is the one to reach for when a problem hands you two velocities and a time and never mentions acceleration. If it will not come to mind under exam pressure, find $a$ first and use a printed equation.
Both. The sheet prints three; the standard textbook set is four. The fourth, $\Delta x = \frac{v_0 + v}{2} t$ , follows from the printed position equation plus $a t = v - v_0$ , and holds only for constant acceleration. For the runner, $\Delta x = 20.0 \ \mathrm{m}$ by either route.

Frequently asked questions

What are the four kinematic equations?

The four standard constant-acceleration equations are $v = v_0 + a t$ , $\Delta x = v_0 t + \frac{1}{2} a t^2$ , $v^2 = v_0^2 + 2 a \Delta x$ , and $\Delta x = \frac{v_0 + v}{2} t$ . The AP Physics 1 equation sheet prints the first three and leaves out the fourth, so on exam day you either recall it or derive it by noting that displacement equals average velocity times time and that, for constant acceleration, average velocity is the plain mean of the starting and ending velocities. All four require the acceleration to be constant over the interval you apply them to.

Which kinematic equation should I use?

Pick the one that does not contain the variable you neither know nor need. Constant-acceleration problems involve five quantities: displacement, initial velocity, final velocity, acceleration, and time. A typical problem gives you three and asks for a fourth, which leaves one quantity unmentioned. Each equation omits exactly one of the five, so the unmentioned quantity names your equation. No time in the problem means use $v^2 = v_0^2 + 2 a \Delta x$ ; no final velocity means use $\Delta x = v_0 t + \frac{1}{2} a t^2$ ; no displacement means use $v = v_0 + a t$ ; no acceleration means use $\Delta x = \frac{v_0 + v}{2} t$ .

What is the kinematic equation for time?

There is no single equation for time; which one you use depends on what else you were given. If you know the two velocities and the acceleration, rearrange $v = v_0 + a t$ to get $t = \frac{v - v_0}{a}$ , which is the quickest route. If you know the displacement and the acceleration but only one velocity, put the numbers into $\Delta x = v_0 t + \frac{1}{2} a t^2$ and solve the resulting quadratic in $t$ , keeping the positive root that matches the physical situation. A dropped object is the simplest case: with $v_0 = 0$ , that reduces to $t = \sqrt{2 \Delta x / g}$ .

Do the kinematic equations work for free fall?

Yes, without modification. Free fall is just constant-acceleration motion along a vertical axis, so all four equations apply with $a$ replaced by the acceleration due to gravity. Choose a positive direction first. With up positive, the acceleration is $-9.8 \ \mathrm{m/s^2}$ for the whole flight, on the way up, at the highest point, and on the way down; with down positive it is $+9.8 \ \mathrm{m/s^2}$ . The most common error is setting the acceleration to zero at the top of a throw because the velocity is zero there. The velocity is zero for an instant; the acceleration never is.

Should I use 9.8 or 10 for g in kinematics problems?

The AP Physics 1 course description states both. Essential knowledge 1.3.A.3 gives the acceleration due to gravity near Earth's surface as approximately $10 \ \mathrm{m/s^2}$ , and a boundary statement in Topic 1.3 says that value will be used for all situations in which a numerical quantity is required for $g$ , while adding that students will not be penalized for correctly using the more precise commonly accepted values of $9.81 \ \mathrm{m/s^2}$ or $9.8 \ \mathrm{m/s^2}$ . The Table of Information prints $9.8 \ \mathrm{m/s^2}$ . Every problem in this set uses 9.8, which typically shifts an answer by about two percent against a value of 10.

How do you find displacement from a velocity-time graph?

Displacement is the area between the velocity curve and the time axis over the interval you care about, counting area below the axis as negative. That is essential knowledge 1.3.A.4.iii in the AP Physics 1 course description. Break the shape into rectangles and trapezoids and add them. The slope of the same graph is the instantaneous acceleration, so one graph answers both kinds of question. Watch the distinction between distance and displacement: if part of the graph dips below the axis, distance adds the magnitudes of all the areas while displacement subtracts the negative ones.

Do these equations still work when the acceleration changes?

No. Every equation in this set assumes the acceleration is constant across the interval you apply it to. If the acceleration varies, the usual fix is to split the motion into intervals where it is constant and solve each in turn, which is exactly what the two-phase problems here do. If it varies continuously, AP Physics 1 will not ask you to compute it: a boundary statement in Topic 1.3 says the course does not expect students to quantitatively analyze nonuniform acceleration, though it does expect you to analyze such situations qualitatively, sketch appropriate graphs of them, and discuss them.

12 Kinematics Practice Problems With Full Solutions

What this set covers

The equations, and which ones the sheet prints

Conventions, and how to work through the set

Practice problems

1. Choose the equation before you calculate

2. Final velocity from an acceleration and a time

3. Displacement from rest

4. Time for a cyclist to slow down

5. Acceleration from two speeds and a distance

6. A dropped wrench

7. Thrown straight up: signs and symmetry

8. Reading a velocity-time graph

9. Two-phase trip: accelerate, then cruise

10. Reaction time plus braking

11. Does a falling object cover 9.8 m in the second second?

12. Three equations or four?

Frequently asked questions