4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A cart moves up a straight ramp, slowing down at a steady rate. It turns around at its highest point and then speeds up at the same steady rate as it rolls back down. The direction up the ramp is positive. Which description matches the cart's velocity–time graph?
Answer and reasoning
AA straight line down to the time axis, then a straight line back up A student who plots the speed rather than the velocity picks this. After the turning point the cart moves down the ramp, the negative direction, so its velocity is negative and grows in size below the axis; the graph does not bounce back up.
BA single straight line that slopes downward and crosses the time axisCorrect The velocity starts positive (up the ramp) and decreases at a steady rate to zero at the turning point, then keeps decreasing at the same rate, becoming more and more negative (down the ramp). One constant slope and a change of sign give one straight line that passes through the time axis.
CA line that rises to a peak and then falls, just as the cart does A student who reads the graph as a picture of the cart going up and coming down picks this. The height of a velocity–time graph is the velocity, which is greatest at the start and decreases steadily: the graph never rises.
DA line down to the time axis, then a flat part along the axis A student who thinks the cart rests for a while at the top picks this. The cart's velocity is zero for only an instant; the same steady change continues through the turning point, so there is no flat part.
A cart moves along a straight track in the +x direction with initial velocity vx0. It has a constant acceleration ax, also in the +x direction. Which expression gives the cart's displacement during the time its velocity increases from vx0 to 3vx0?
Answer and reasoning
A6vx0²/ax A student who multiplies the final velocity by the time, 3vx0 × (2vx0/ax), picks this. The cart is slower than 3vx0 for the whole interval, so this overestimates the displacement; the velocity changes, so one value cannot stand for the whole interval.
B8vx0²/ax A student who drops the 2 from vx² = vx0² + 2ax(x − x0) gets 9vx0² − vx0² = axΔx and picks this. The factor 2 belongs in the equation; with it, Δx = 8vx0²/(2ax) = 4vx0²/ax.
C2vx0²/ax A student who finds the time, t = 2vx0/ax, and then uses Δx = (1/2)ax t² as if the cart started from rest picks this. The vx0 t term, another 2vx0²/ax, is missing.
D4vx0²/axCorrect Using vx² = vx0² + 2ax(x − x0): (3vx0)² = vx0² + 2axΔx, so 2axΔx = 8vx0² and Δx = 4vx0²/ax.
Working vx² = vx0² + 2ax(x − x0) with vx = 3vx0: 9vx0² − vx0² = 2axΔx, so Δx = 8vx0²/(2ax) = 4vx0²/ax. Check with time: t = (3vx0 − vx0)/ax = 2vx0/ax; Δx = vx0 t + (1/2)ax t² = 2vx0²/ax + 2vx0²/ax = 4vx0²/ax.
A ball is thrown vertically upward, reaches its highest point, and falls back down. Air resistance is negligible. Which statement correctly compares the ball's acceleration at different stages of its flight?
Answer and reasoning
AIt is upward while the ball rises and downward while the ball falls back down. A student who thinks acceleration points in the direction of motion picks this. While rising, the ball slows down, so its velocity change is downward: the acceleration is downward on the way up as well.
BIt is downward and the same size while rising, at the top, and while falling.Correct Near Earth's surface the only influence on the ball is gravity, so its acceleration is g, downward and constant, at every stage. At the top the velocity is momentarily zero but is still changing from upward to downward.
CIt is downward while rising and while falling, but it is zero at the highest point. A student who thinks zero velocity means zero acceleration picks this. At the top the velocity is zero for only an instant; it is still changing at 10 m/s each second, so the acceleration there is still g downward.
DIt is downward throughout, but it is smallest near the top, where the ball is slowest. A student who links the size of the acceleration to the speed picks this. The velocity changes by the same amount each second at every stage, however slowly the ball is moving, so the acceleration has the same size throughout.
The graph shows the position x of a cart moving along a straight track as a function of time t. The dashed line is tangent to the curve at point P, where t = 3.0 s. What is the cart's instantaneous velocity at t = 3.0 s?
Answer and reasoning
A2.5 m/s A student who divides the position at P by the time, 7.5 m ÷ 3.0 s, picks this. That is the slope of the line from the origin to P, the AVERAGE velocity over the first 3.0 s; the curve is getting steeper, so the velocity at 3.0 s is greater than that average.
B7.5 m/s A student who reads the height of the graph at P as the velocity picks this. The height of a position–time graph is the position, 7.5 m; the velocity is the slope of the tangent.
C0.2 m/s A student who divides the change in time by the change in position, 3.0 s ÷ 15 m, picks this. Slope is the change in the vertical-axis quantity divided by the change in the horizontal-axis quantity, Δx/Δt, in m/s.
D5.0 m/sCorrect The instantaneous velocity is the slope of the tangent at P. Using the two marked points on the tangent, (15 − 0) m ÷ (4.5 − 1.5) s = 5.0 m/s.
Working Tangent passes through (1.5 s, 0) and (4.5 s, 15 m): vx = Δx/Δt = (15 − 0) m ÷ (4.5 − 1.5) s = 5.0 m/s. (The curve is x = (5/6)t², constant acceleration 5/3 m/s²; the average velocity 0–3.0 s is 7.5 m ÷ 3.0 s = 2.5 m/s.)
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.3.A.1 Motion diagram Fix
Motion diagram
A diagram that shows an object's position at a series of instants separated by equal time intervals, drawn as dots (or images) along the line of motion. Equal spacing shows constant velocity; spacing that grows or shrinks shows that the object is speeding up or slowing down, and the change in spacing from one interval to the next shows the direction of the acceleration.
Representations of motion
The same motion can be described by a motion diagram, a figure, a graph of position, velocity or acceleration against time, an equation such as x = x0 + vx0 t + (1/2)ax t², or a narrative description in words. Each representation must agree with the others: for example, a narrative of an object that slows down, reverses and speeds up again corresponds to a velocity–time graph that crosses the time axis.
Students often think An object that is slowing down is not accelerating: 'acceleration' means speeding up, so a slowing object has zero acceleration. In fact Yes. In physics, acceleration is any change in velocity: speeding up, slowing down or changing direction. A slowing object has an acceleration directed opposite to its velocity.
Students often think An object that reverses direction stays at rest for a short time at the turning point before it starts to move back. In fact No. Its velocity is zero for only an instant. The acceleration is still nonzero, so the velocity keeps changing and the object moves straight back. On a velocity–time graph the line passes through the time axis without a flat part.
1.3.A.2 Constant (uniform) acceleration Fix
Constant (uniform) acceleration
Motion in which the velocity changes by equal amounts in equal time intervals, so the acceleration ax has the same value at every instant. Its velocity–time graph is a single straight line and its acceleration–time graph is a horizontal line. SI unit of acceleration: m/s².
Kinematic equations
For constant acceleration in one dimension: vx = vx0 + ax t; x = x0 + vx0 t + (1/2)ax t²; vx² = vx0² + 2ax(x − x0). Here x0 and vx0 are the position and velocity at t = 0. Each equation leaves out one quantity (x, vx or t respectively). They apply along any single axis, including the vertical, and only while the acceleration is constant; they remain valid through an instant at which the velocity is zero.
Sign convention in one dimension
One direction along the line of motion is chosen as positive. Position, displacement, velocity and acceleration are then written as signed components (x, Δx, vx, ax): the sign gives the direction and must be applied to every quantity in the kinematic equations, including ax = −g when upward is positive.
Students often think The displacement of an accelerating object is its velocity multiplied by the time, Δx = vx t, using the initial or the final velocity. In fact Not with a single instantaneous velocity. For constant acceleration use x = x0 + vx0 t + (1/2)ax t², or the average of the initial and final velocities multiplied by the time. Δx = vx t holds only when the velocity is constant.
Students often think The initial velocity can be left out: formulas and graph areas can be used as if the object started from rest, so x − x0 = (1/2)ax t² and vx = ax t, and the area under an acceleration–time graph is the final velocit… In fact Only if the object starts from rest. Otherwise the vx0 terms must be kept, and the area under an acceleration–time graph gives the change in velocity, which must be added to the initial velocity.
1.3.A.3 Acceleration due to gravity, g Fix
Acceleration due to gravity, g
Near Earth's surface, an object moving only under the influence of gravity (air resistance negligible) has a constant acceleration directed downward, of magnitude g ≈ 10 m/s² in AP Physics 1. It is the same whether the object is rising, at its highest point or falling, and whatever its mass. With upward positive, ay = −g.
Free fall and the highest point
An object in free fall has acceleration g downward throughout its flight. At the highest point of a vertical throw its velocity is zero for an instant, but the velocity is still changing (from upward to downward), so the acceleration there is still g downward.
Students often think If an object's velocity is zero at an instant, its acceleration must also be zero at that instant. In fact Not necessarily. Acceleration is the rate of change of velocity. At the top of a vertical throw, or wherever an object reverses, the velocity is zero for an instant but is still changing, so the acceleration is not zero.
Students often think An object's acceleration points in its direction of motion, so an object moving upward (or to the right) has an upward (or rightward) acceleration, and the acceleration reverses when the object reverses. In fact No. The velocity gives the direction of motion; the acceleration gives the direction in which the velocity is changing, which is opposite to the motion when the object slows down. A ball thrown upward moves up while its acceleration is downward.
1.3.A.4 Position–time, velocity–time and acceleration–time graphs Fix
Position–time, velocity–time and acceleration–time graphs
Graphs of x, vx and ax against t describe the same motion. The slope of each graph at an instant gives the value of the next quantity (x → vx → ax), and the area under each of the last two over an interval gives the change in the previous one (ax → Δvx, vx → Δx). The height of a graph is simply the value of the quantity plotted.
Instantaneous velocity
The rate of change of position at one instant, equal to the slope of the line tangent to the position–time graph at that instant. It is a signed component along the axis: positive slope means motion in the positive direction, zero slope means momentarily at rest. Its magnitude is the instantaneous speed. SI unit: m/s.
Tangent line and slope
A tangent line touches a curve at one point and has the same steepness as the curve there. Its slope is Δ(vertical quantity)/Δ(horizontal quantity), found from two well-separated points on the tangent line itself. For a straight-line graph the tangent at every point is the line itself.
Instantaneous acceleration
The rate of change of velocity at one instant, equal to the slope of the line tangent to the velocity–time graph at that instant. It can be nonzero when the velocity is zero, and its sign gives its direction, not whether the object is speeding up. SI unit: m/s².
Speeding up and slowing down
An object speeds up when its velocity and acceleration have the same sign (same direction) and slows down when they have opposite signs. A decreasing but still positive acceleration, with positive velocity, means the object is still speeding up, only at a smaller rate.
Displacement from a velocity–time graph
The displacement Δx during an interval equals the area bounded by the velocity–time graph and the time axis over that interval. Area above the axis counts as positive displacement and area below as negative; adding the sizes of all the areas instead gives the distance traveled. SI unit: m.
Change in velocity from an acceleration–time graph
The change in velocity Δvx during an interval equals the area bounded by the acceleration–time graph and the time axis, with area below the axis negative. The velocity at the end of the interval is the initial velocity plus this change: vx = vx0 + Δvx. SI unit: m/s.
Nonuniform acceleration (qualitative)
Acceleration that changes with time. It shows as a curved velocity–time graph (its steepness changing) and a position–time graph whose curvature changes. In AP Physics 1 it is analyzed qualitatively: which part is steeper, where the slope is zero, how the sign of the slope changes.
Students often think A motion graph is a picture of the motion: a line that rises and then falls shows an object going up and coming back down, and a level line shows an object at rest, whatever quantity is plotted. In fact No. A motion graph shows how one quantity changes with time, and its shape need not resemble the path. A cart that rolls up a ramp and back down with constant acceleration has a velocity–time graph that is one straight line; a level line on a velocity–time graph means constant velocity, not rest.
Students often think The value of a quantity at an instant can be read as the height of whatever motion graph is given: the height of a position–time graph gives the velocity, the height of a velocity–time graph gives the position or the ac… In fact No. The height of a graph is the value of the quantity plotted on its vertical axis. Velocity is the slope of a position–time graph, acceleration is the slope of a velocity–time graph, and a velocity is found from an acceleration–time graph only through the area (a change in velocity).
10 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 10
The motion diagram shows the position of a cart at six instants separated by equal time intervals, as the cart moves to the right along a straight track. The dots are numbered in time order. Which claim about the cart's acceleration is supported by the diagram?
Answer and reasoning
AIt points to the right, since the cart moves to the right during the whole time shown. A student who thinks acceleration points in the direction of motion picks this. The cart does move to the right, but it covers less distance in each interval, so its velocity is decreasing: the velocity change, and so the acceleration, points to the left.
BIt is zero, since the cart is slowing down rather than speeding up as it goes. A student who thinks 'acceleration' means only speeding up picks this. Slowing down is a change in velocity too, so the cart has a nonzero acceleration, directed opposite to its motion (to the left).
CIt points to the left, since the cart moves a shorter distance in each interval.Correct The gaps between successive dots are 5 m, 4 m, 3 m, 2 m and 1 m. Equal time intervals with shrinking gaps mean the cart's velocity to the right is decreasing, so the change in velocity, and hence the acceleration, points to the left. (Each gap is 1 m shorter than the one before, so the acceleration is constant.)
DIt is decreasing in size, since the gaps between the dots get smaller and smaller. A student who links the size of the acceleration to the speed picks this. Smaller gaps show a smaller VELOCITY. Each gap is exactly 1 m shorter than the one before, so the velocity drops by the same amount in every interval: the acceleration is constant, not decreasing.
Working Gaps: 5, 4, 3, 2, 1 m in equal time intervals. The displacement per interval (average velocity) falls by 1 m each interval, so the velocity to the right decreases at a steady rate: acceleration constant and directed to the left.
A ball is thrown vertically upward with initial velocity vy0. Air resistance is negligible. Taking upward as positive, which expression gives the ball's vertical displacement Δy from the launch point to its highest point?
Answer and reasoning
Avy0²/(2g)Correct At the highest point vy = 0, and the acceleration is ay = −g throughout. From vy² = vy0² + 2ayΔy: 0 = vy0² − 2gΔy, so Δy = vy0²/(2g).
Bvy0²/g A student who multiplies the launch velocity by the time to reach the top, vy0 × (vy0/g), picks this. The ball slows down all the way up, so its average velocity is only vy0/2 and the displacement is half this value.
C3vy0²/(2g) A student who takes the acceleration to be upward while the ball rises picks this: finding the time to the top as vy0/g and then using Δy = vy0 t + (1/2)ay t² with ay = +g gives vy0²/g + vy0²/(2g) = 3vy0²/(2g). The acceleration is downward (ay = −g) even while the ball moves up, which makes the second term −vy0²/(2g) and gives vy0²/(2g).
Dvy0/(2g) A student who forgets to square vy0 when substituting picks this. Its units are (m/s)/(m/s²) = s, a time, not a displacement.
Working Upward positive, ay = −g, vy = 0 at the top. vy² = vy0² + 2ayΔy → 0 = vy0² − 2gΔy → Δy = vy0²/(2g). Units: (m²/s²)/(m/s²) = m. Distractor 3vy0²/(2g): t = vy0/g, ay = +g in Δy = vy0 t + (1/2)ay t².
A cart is released from rest and rolls along a straight track with constant acceleration. It travels a distance d during the first time interval T after its release. The distance it travels during the next interval T, from T to 2T, is how many times d?
Answer and reasoning
A1 A student who thinks constant acceleration means equal distances in equal times picks this. That is true of constant VELOCITY; here the velocity grows, so each interval covers more distance than the one before.
B2 A student who multiplies the velocity at time T, axT, by the interval T picks this, getting axT² = 2d. The cart keeps speeding up during the second interval, so it travels farther than that.
C4 A student who gives the cart's distance from the start at time 2T picks this. That, 4d, includes the first interval; the distance traveled from T to 2T is 4d − d.
D3Correct From rest, the distance traveled is proportional to t²: by time T the cart has gone d = (1/2)axT², and by 2T it has gone (1/2)ax(2T)² = 4d. During the second interval it travels 4d − d = 3d.
Working x − x0 = (1/2)ax t² from rest. At T: d = (1/2)axT². At 2T: (1/2)ax(2T)² = 4 × (1/2)axT² = 4d. Second interval: 4d − d = 3d, so the answer is 3.
An object falls freely near Earth's surface. Air resistance is negligible. Use g = 10 m/s². Which statement about the object's motion is correct?
Answer and reasoning
AIt falls through a distance of 10 m during each second of its fall. A student who reads 10 m/s² as '10 m every second' picks this. The unit means the VELOCITY changes by 10 m/s each second. The distances fallen in successive seconds from rest are 5 m, 15 m, 25 m …, not a steady 10 m.
BIts acceleration is greater than 10 m/s² if it has a large mass. A student who thinks heavier objects fall faster picks this. With air resistance negligible, every object near Earth's surface falls with the same acceleration, g, whatever its mass.
CIts velocity changes by 10 m/s downward during each second of its fall.Correct An acceleration of 10 m/s² downward means the velocity changes by 10 m/s downward in every second: from rest, 0, 10, 20, 30 m/s downward at 0, 1, 2, 3 s.
DIts acceleration increases as its speed increases during the fall. A student who links the size of the acceleration to the speed picks this. The speed does increase, but by the same 10 m/s every second, so the acceleration stays constant at 10 m/s².
Working a = g = 10 m/s² downward: Δv = aΔt = 10 m/s per 1 s. From rest, distances in successive 1 s intervals: (1/2)(10)(1)² = 5 m, (1/2)(10)(2)² − 5 = 15 m, 45 − 20 = 25 m.
The graph shows the velocity v of an object moving along a straight line as a function of time t. Which statement about the object's motion from t = 0 to t = 4 s is correct?
Answer and reasoning
AIts position is least at t = 2 s, the instant when its velocity is zero.Correct From 0 to 2 s the velocity is negative, so the object moves in the negative direction and its position decreases. After 2 s the velocity is positive and the position increases. The position is therefore least at t = 2 s, when the object reverses.
BIts position is least at t = 0, the instant when its velocity is most negative. A student who reads the height of the velocity graph as the position picks this. At t = 0 the object is moving in the negative direction, so its position is still decreasing; it keeps decreasing until the velocity reaches zero at t = 2 s.
CIts speed increases throughout, since its acceleration is positive. A student who thinks a positive acceleration always means speeding up picks this. From 0 to 2 s the velocity is negative and the acceleration positive, so the object slows down (its speed falls from 4 m/s to 0); it speeds up only after 2 s.
DIts acceleration is zero at t = 2 s, the instant when its velocity is zero. A student who thinks zero velocity means zero acceleration picks this. The graph is a straight line with the same slope, +2 m/s², at t = 2 s as at every other time; only the velocity is zero there.
Working Slope = (4 − (−4)) m/s ÷ 4 s = +2 m/s², constant. v < 0 for 0 < t < 2 s (moving in −x, position decreasing, slowing), v = 0 at t = 2 s, v > 0 afterwards (moving in +x, speeding up). Displacement 0–2 s = −4 m, 2–4 s = +4 m, so x is least at t = 2 s.
The graph shows the position x of an object moving along a straight line as a function of time t. Points A, B and C are on the curve; the curve is horizontal at B. Which ranks the object's speeds at A, B and C, from greatest to least?
Answer and reasoning
AB > C > A A student who ranks the heights of the points (8 m, 6 m, 1.5 m) picks this. The height of a position–time graph is the position; B is the farthest point from the origin, but the curve is flat there, so the object is momentarily at rest.
BA > C > BCorrect Speed is the size of the slope of the tangent to the position–time graph. The curve is steepest at A (a slope of about 3 m/s), less steep at C (sloping down, about 1.5 m/s), and horizontal at B (zero), so A > C > B.
CA > B > C A student who ranks the signed velocities picks this: positive at A, zero at B, negative at C. The question asks for speeds, the magnitudes of the slopes; the object moves faster at C (in the negative direction) than at B, where it is at rest.
DB > A > C A student who takes x/t at each point as the speed (1.5 m/s at A, 2.0 m/s at B, about 0.9 m/s at C) picks this. x/t is the slope of a line from the origin, an average velocity; the speed at an instant is the slope of the tangent, which is zero at B.
Working Tangent slopes: A about +3 m/s (steep, rising), B 0 (horizontal), C about −1.5 m/s (falling, less steep than at A). Speeds: A ≈ 3 m/s > C ≈ 1.5 m/s > B = 0.
The graph shows the velocity v of an object moving along a straight line as a function of time t. A student claims that the object's acceleration is zero at t = 3 s, because the object is at rest at that instant. Which statement correctly evaluates the claim?
Answer and reasoning
ACorrect: the velocity is zero at t = 3 s, so the velocity is not changing then. A student who thinks zero velocity means zero acceleration picks this. The velocity is zero at t = 3 s, but it is changing: it goes from positive just before 3 s to negative just after, at a steady −4 m/s².
BIncorrect: the acceleration changes from positive to negative at the instant t = 3 s. A student who thinks acceleration points in the direction of motion picks this, because the object moves in the positive direction before 3 s and the negative direction after. The slope, and so the acceleration, is negative throughout.
CIncorrect: the graph has the same slope at t = 3 s as at every other time.Correct Acceleration is the slope of the velocity–time graph. The graph is one straight line, so its slope, (−4 − 12) m/s ÷ 4 s = −4 m/s², is the same at every instant, including t = 3 s where the velocity passes through zero.
DIncorrect: the acceleration is negative before t = 3 s and positive after it. A student who thinks a positive acceleration means speeding up picks this, because the object slows before 3 s and speeds up after. After 3 s it speeds up in the negative direction; the slope is still negative, so the acceleration is still −4 m/s².
Working Slope = (−4 − 12) m/s ÷ (4 − 0) s = −4 m/s² at every instant. v = 12 − 4t = 0 at t = 3 s. So a = −4 m/s² at t = 3 s: the claim is incorrect.
The graph shows the velocity v of a car as a function of time t as the car pulls away from rest along a straight, level road. Which statement describes the car's motion during the time shown?
Answer and reasoning
AIts speed rises to a steady value as its acceleration decreases to zero.Correct The velocity (height) rises and then levels off, so the speed increases to a steady value. The slope, which is the acceleration, is greatest at the start and decreases to zero where the graph becomes horizontal.
BIts speed and its acceleration both increase during the time shown. A student who links the size of the acceleration to the speed picks this. The speed does increase, but the slope of the graph decreases: the car gains speed ever more slowly, so its acceleration decreases.
CIt slows down, since its acceleration decreases as time goes on. A student who thinks a decreasing acceleration means slowing down picks this. The acceleration does decrease, but it stays in the direction of motion until it reaches zero, so the car keeps speeding up; the velocity never falls.
DIt comes to rest at the instant when the graph becomes a horizontal line. A student who reads a level line as 'not moving', as on a position–time graph, picks this. On a velocity–time graph a level line well above the time axis means a constant, nonzero velocity: the car keeps moving at a steady speed.
The graph shows the velocity v of a cart moving along a straight track as a function of time t. What is the cart's displacement from t = 0 to t = 3 s?
Answer and reasoning
A+5 m A student who adds both areas as positive, 4 m + 1 m, picks this. That is the distance traveled. The cart moves back 1 m after t = 2 s, so its displacement is 4 m − 1 m.
B−6 m A student who multiplies the final velocity by the time, (−2 m/s)(3 s), picks this. The velocity changes throughout the interval, so no single value of it can be multiplied by the time; the area under the graph is needed.
C−2 m A student who finds the slope of the graph, (−2 − 4) m/s ÷ 3 s = −2, and reports it as the displacement picks this. The slope is the acceleration, −2 m/s²; the displacement is the area.
D+3 mCorrect Displacement is the area between the graph and the time axis, counting area below the axis as negative. From 0 to 2 s: (1/2)(2 s)(4 m/s) = +4 m. From 2 to 3 s: (1/2)(1 s)(−2 m/s) = −1 m. Total +4 m − 1 m = +3 m.
Working Line from (0, +4 m/s) to (3 s, −2 m/s); slope −2 m/s², v = 0 at t = 2 s. Area 0–2 s = (1/2)(2)(4) = +4 m; area 2–3 s = (1/2)(1)(−2) = −1 m. Δx = +3 m. (Check: average velocity (4 + (−2))/2 = 1 m/s × 3 s = 3 m.)
The graph shows the acceleration a of a cart moving along a straight track as a function of time t. At t = 0 the cart's velocity is +2 m/s. What is the cart's velocity at t = 4 s?
Answer and reasoning
A+5 m/s A student who takes the area under the graph to be the final velocity picks this. The area, +5 m/s, is the CHANGE in velocity; the cart already had +2 m/s at t = 0, so its velocity at 4 s is +7 m/s.
B+9 m/s A student who adds the area below the axis as positive, 2 + 6 + 1, picks this. Between 3 s and 4 s the acceleration is negative, so that area reduces the velocity by 1 m/s.
C+7 m/sCorrect The change in velocity is the area under the acceleration–time graph: (+2 m/s²)(3 s) = +6 m/s from 0 to 3 s, and (−1 m/s²)(1 s) = −1 m/s from 3 to 4 s, a total of +5 m/s. Adding the initial velocity: +2 m/s + 5 m/s = +7 m/s.
D−1 m/s A student who reads the height of the graph at t = 4 s as the velocity picks this. The graph shows acceleration, −1 m/s² at that time; the velocity comes from the initial velocity plus the area under the graph.
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
