B. Kinematics · Kinematics: Motion in One and Two DimensionsAcceleration and Velocity-Time Graphs
B2.1 B2.2 B3.1 B3.2 B2.7About 75 minutes
Learning goals- I can define acceleration and calculate average acceleration with direction.
- I can tell from the signs of velocity and acceleration whether an object is speeding up or slowing down.
- I can find acceleration as the slope of a velocity-time graph.
- I can find displacement as the area between a velocity-time graph and the time axis.
Words to know
acceleration (a)
The rate of change of velocity: aₐᵥ = Δv ÷ Δt = (v₂ − v₁) ÷ Δt. A vector; SI unit m/s².
m/s²
Metres per second per second: how many m/s the velocity changes each second.
velocity-time graph (v–t graph)
A graph with time on the horizontal axis and velocity on the vertical axis.
uniformly accelerated motion
Motion with constant acceleration. Its v–t graph is a straight line.
instantaneous acceleration
The acceleration at one instant: the slope of the tangent to a v–t graph.
The feeling of motion
Sit in a car (as a passenger!) at a steady 100 km/h on a smooth highway and close your eyes. You can barely tell you are moving. But when the car pulls away from a stop sign, brakes hard, or swings around a curve, you feel it right away. What you feel is not velocity. It is a change in velocity: acceleration.
What acceleration means
Acceleration is the rate at which velocity changes:
aₐᵥ = Δv ÷ Δt = (v₂ − v₁) ÷ Δt
where v₁ is the initial velocity and v₂ is the final velocity. The unit is (m/s) ÷ s = m/s². An acceleration of 3.0 m/s² [E] means the velocity changes by 3.0 m/s [E] every second.
Acceleration is a vector. Its direction is the direction of the change in velocity, which is not always the direction of motion. Velocity can change in three ways: the object speeds up, slows down, or changes direction. All three are acceleration. (Turning at a constant speed is covered in Grade 12 Physics, SPH4U; in this course we focus on straight-line changes.)
Worked exampleExample 1: Speeding up
A car starts from rest at a green light and reaches 27 m/s [E] in 9.0 s. Find its average acceleration.
- East positive. v₁ = 0 ("from rest"), v₂ = +27 m/s, Δt = 9.0 s.
- aₐᵥ = (v₂ − v₁) ÷ Δt = (27 − 0) ÷ 9.0 = +3.0 m/s².
- So aₐᵥ = 3.0 m/s² [E]: each second the car gains 3.0 m/s of eastward velocity.
Worked exampleExample 2: Slowing down
A cyclist moving at 8.0 m/s [E] brakes and slows to 2.0 m/s [E] in 3.0 s. Find her average acceleration.
- East positive. v₁ = +8.0 m/s, v₂ = +2.0 m/s, Δt = 3.0 s.
- aₐᵥ = (2.0 − 8.0) ÷ 3.0 = −6.0 ÷ 3.0 = −2.0 m/s².
- The negative sign means the acceleration points west: aₐᵥ = 2.0 m/s² [W]. The cyclist is still moving east, but her acceleration points the opposite way, so she slows down.
Speeding up or slowing down?
A negative acceleration does not always mean slowing down. It means the acceleration points in the negative direction. What matters is how the acceleration compares with the velocity:
- Same direction (same signs) → the object speeds up.
- Opposite directions (opposite signs) → the object slows down.
Many people say "deceleration" for slowing down. Physicists avoid the word because it hides the direction. Say "an acceleration of 2.0 m/s² [W]" instead.
Velocity-time graphs
A velocity-time (v–t) graph has time on the horizontal axis and velocity on the vertical axis. It holds two kinds of information:
- Slope = acceleration. Slope = Δv ÷ Δt = a. A horizontal line means constant velocity (a = 0). A straight sloping line means constant acceleration (uniformly accelerated motion). A curved line means the acceleration is changing; the slope of a tangent gives the instantaneous acceleration.
- Area = displacement. The area between the graph and the time axis equals Δd. Why? For constant velocity, Δd = vΔt, which is height × width of a rectangle. The same idea works for any shape, split into rectangles and triangles. Area above the time axis counts as positive displacement; area below counts as negative.
Where a v–t graph crosses the time axis, the velocity is zero for an instant and the object changes direction.
Worked exampleExample 3: Reading a v–t graph
Use the graph above. Find (a) the acceleration in each segment and (b) the total displacement and average velocity for the 13 s.
- (a) 0 to 4 s: a = (12 − 0) ÷ 4 = 3.0 m/s² [E]. 4 to 10 s: the line is horizontal, so a = 0. 10 to 13 s: a = (0 − 12) ÷ (13 − 10) = −4.0 m/s², or 4.0 m/s² [W] (slowing down while moving east).
- (b) Triangle: ½ × 4 s × 12 m/s = 24 m. Rectangle: 6 s × 12 m/s = 72 m. Triangle: ½ × 3 s × 12 m/s = 18 m.
- Total Δd = 24 + 72 + 18 = 114 m [E].
- vₐᵥ = 114 m ÷ 13 s ≈ 8.8 m/s [E].
a = 3.0 m/s² [E], 0, 4.0 m/s² [W]; Δd = 114 m [E]; vₐᵥ ≈ 8.8 m/s [E].
Worked exampleExample 4: Turning around on a ramp
A ball is rolled up a long ramp at 3.0 m/s. It slows, stops, and rolls back down. After 4.0 s it is moving at 1.0 m/s down the ramp. Take up the ramp as positive. Find (a) the acceleration, (b) when the ball stops, and (c) its displacement and distance travelled in the 4.0 s.
- (a) v₁ = +3.0 m/s and v₂ = −1.0 m/s. a = (−1.0 − 3.0) ÷ 4.0 = −1.0 m/s², or 1.0 m/s² [down the ramp]. The acceleration is the same while the ball goes up and comes down.
- (b) The ball stops when v = 0. Starting at 3.0 m/s and losing 1.0 m/s each second, that happens at t = 3.0 s, where the graph crosses the time axis.
- (c) Area above the axis: ½ × 3.0 × 3.0 = +4.5 m. Area below: ½ × 1.0 × (−1.0) = −0.5 m.
- Displacement = 4.5 + (−0.5) = 4.0 m [up the ramp]. Distance = 4.5 + 0.5 = 5.0 m.
(a) 1.0 m/s² [down the ramp]; (b) at 3.0 s; (c) Δd = 4.0 m [up the ramp], distance = 5.0 m.
Example 4 as a v–t graph. The straight line means constant acceleration (slope −1.0 m/s²). The ball turns around where the line crosses the time axis. Area above the axis is positive displacement; area below is negative.Just for funMy driving instructor says I accelerate very smoothly.
She has also asked me to stop calling the brake "the negative-acceleration pedal"... because if we're reversing, it's the positive-acceleration pedal.
Hands-on activityA v–t graph from the passenger seat
Grown-up note: the student is the passenger only. The driver drives normally and follows all traffic laws; nobody touches a phone while driving.
You need: a stopwatch (or a phone used only by the passenger), a notebook and graph paper.
- When the car pulls away from a stop sign or red light on a quiet road, start timing. Every 2 s, read the speedometer (km/h) aloud and write it down, until the car reaches a steady speed.
- Convert each reading to m/s (÷ 3.6).
- Plot a v–t graph. Draw a best-fit line or curve.
- Find the average acceleration from the slope (m/s²).
- Estimate the displacement during your data by finding the area under the graph (count squares or use triangles and rectangles).
What to look for: a typical gentle start from rest reaches 50 km/h (about 14 m/s) in roughly 7 to 10 s, so the acceleration is roughly 1.5 to 2 m/s². The graph is usually steepest at the start and flattens as the car reaches its cruising speed, which shows the acceleration was not constant.
Your turn
1A skateboarder speeds up from 2.0 m/s [S] to 6.5 m/s [S] in 3.0 s. Find the average acceleration.
South positive: a = (6.5 − 2.0) ÷ 3.0 = 1.5 m/s² [S].
2A train slows from 24 m/s [W] to 8.0 m/s [W] in 20.0 s. Find its average acceleration.
West positive: a = (8.0 − 24) ÷ 20.0 = −0.80 m/s², so 0.80 m/s² [E].
3An object has v = −5 m/s and a = +2 m/s². Is it speeding up or slowing down?
Slowing down: the velocity and acceleration have opposite signs (opposite directions).
4A v–t graph is a horizontal line at 15 m/s from t = 0 to t = 8.0 s. Find the acceleration and the displacement.
Slope 0, so a = 0. Area = 15 × 8.0 = 120 m in the positive direction.
5A v–t graph is a straight line from (0 s, 4.0 m/s) to (5.0 s, 14.0 m/s). Find the acceleration and the displacement.
a = (14.0 − 4.0) ÷ 5.0 = 2.0 m/s². Area (trapezoid) = ½(4.0 + 14.0)(5.0) = 45 m.
6What does it mean when a v–t graph crosses the time axis?
The velocity is zero for an instant and the object changes direction.
7A car moving at 20 m/s [N] has an acceleration of 4.0 m/s² [S]. What is its velocity 3.0 s later?
North positive: Δv = aΔt = (−4.0)(3.0) = −12 m/s, so v₂ = 20 − 12 = 8.0 m/s [N].
More practice: warm-up
1What is the SI unit of acceleration?
m/s² (metres per second squared).
2What does the slope of a v–t graph give?
3What does the area under a v–t graph give?
4A bus goes from 0 to 12 m/s in 6.0 s. Magnitude of acceleration?
5Is a car moving at a steady 90 km/h in a straight line accelerating?
No. Its velocity is not changing, so a = 0.
More practice: on your own
1A hockey puck hits the boards at 12 m/s [E] and rebounds at 8.0 m/s [W]. The contact lasts 0.040 s. Find the average acceleration.
East positive: a = (−8.0 − 12) ÷ 0.040 = −20 ÷ 0.040 = −500 m/s², so 5.0 × 10² m/s² [W]. Changing direction gives a large Δv.
2A v–t graph goes in straight segments through (0 s, 0), (2.0 s, 6.0 m/s), (5.0 s, 6.0 m/s) and (6.0 s, 0). Find the total displacement.
Triangle ½(2.0)(6.0) = 6.0 m; rectangle (3.0)(6.0) = 18 m; triangle ½(1.0)(6.0) = 3.0 m. Total = 27 m.
3For the same graph, find the acceleration in the last segment and say whether the object is speeding up or slowing down.
a = (0 − 6.0) ÷ (6.0 − 5.0) = −6.0 m/s². The velocity is positive and the acceleration negative, so it is slowing down.
4A straight v–t line goes from (0 s, +6.0 m/s) to (6.0 s, −3.0 m/s). Find the acceleration, the time it turns around, and the displacement.
a = (−3.0 − 6.0) ÷ 6.0 = −1.5 m/s². v = 0 when 6.0 − 1.5t = 0, t = 4.0 s. Areas: ½(4.0)(6.0) = +12 m and ½(2.0)(−3.0) = −3.0 m. Δd = +9.0 m.
5Explain why a car going around a curve at a steady 50 km/h is accelerating.
Acceleration is a change in velocity, and velocity includes direction. On a curve the direction keeps changing, so the velocity changes and the car accelerates, even though its speed is constant.
Challenge
1An object starts at 10 m/s [E] and has a constant acceleration of 2.5 m/s² [W]. Find its velocity after 6.0 s and its displacement over the 6.0 s using a v–t graph.
v₂ = 10 + (−2.5)(6.0) = −5.0 m/s = 5.0 m/s [W]. It stops at t = 10 ÷ 2.5 = 4.0 s. Area above: ½(4.0)(10) = 20 m; below: ½(2.0)(−5.0) = −5.0 m. Δd = 15 m [E] (distance 25 m).
2On a curved v–t graph, the tangent at t = 3.0 s passes through (1.0 s, 2.0 m/s) and (5.0 s, 10.0 m/s). What is the instantaneous acceleration at 3.0 s?
a = (10.0 − 2.0) ÷ (5.0 − 1.0) = 2.0 m/s². The slope of the tangent to a v–t graph is the instantaneous acceleration.
3A ball is thrown straight up. At the very top, its velocity is zero. Is its acceleration also zero there? Use a v–t graph to explain.
No. The v–t graph is a straight line sloping down through the time axis. At the top the line crosses v = 0, but its slope (the acceleration) is the same as everywhere else, not zero. If the acceleration were zero at the top, the ball would stay there. (You will use this in the free-fall lesson.)
Quiz
1. Average acceleration is defined as:
2. A runner slows from 6.0 m/s [N] to 2.0 m/s [N] in 2.0 s. Her acceleration is:
3. An object has a negative velocity and a negative acceleration. It is:
4. On a v–t graph, the area between the line and the time axis gives:
5. A v–t graph is a straight line from (0 s, 0 m/s) to (10 s, 20 m/s). The displacement is:
6. A horizontal line on a v–t graph above the time axis means the object:
Big idea: Acceleration is the rate of change of velocity, a vector (B2.1, B3.2). On a v–t graph, the slope is acceleration and the area is displacement (B2.2). Students also learn the difference between average, instantaneous and constant acceleration (B3.1).
How to help: Ask, "Is the velocity changing? Then there is acceleration." Point out acceleration in daily life: pulling away from a stop, braking, a bus going around a corner.
Common difficulties:
- Thinking negative acceleration always means slowing down. It depends on the direction of the velocity.
- Confusing d–t and v–t graphs (a horizontal line means at rest on one and constant velocity on the other). Have your teen check the vertical-axis label first.
- Forgetting that area below the time axis is negative displacement.
- Writing m/s instead of m/s² for acceleration.