Is acceleration zero at the top of a throw?
No. At the very top, the ball's velocity is zero for one instant, but its acceleration is still g ≈ 9.8 m/s², pointing down.
Acceleration is how fast velocity is changing, not how big it is. At the top the velocity is changing from "going up" to "coming down", at the same rate as at every other moment of the flight. If the acceleration really were zero there, the ball would hang in the air forever.
See it: throw a ball
Drag the slider or press play. Watch the blue velocity arrow shrink, vanish at the top, and grow again downwards, while the green acceleration arrow never changes.
The ball
Its velocity–time graph
Going up: velocity is positive and shrinking. The line slopes down the whole time: that slope is the acceleration.
Ball thrown straight up at 19.6 m/s, air resistance ignored, g = 9.8 m/s². It reaches the top after 2 s and lands after 4 s.
What acceleration actually is
Galileo noticed that a falling object gains the same velocity every second (about 9.8 m/s each second, ignoring air) but not the same velocity every metre, because it covers more metres each second. That is why acceleration is defined per second, not per metre:
average acceleration = Δv / Δt (unit: m/s²)- Average acceleration. A car goes from 10 m/s at t = 2 s to 20 m/s at t = 7 s. Δv = 10 m/s over Δt = 5 s, so a = 2 m/s²: on average the velocity grows by 2 m/s every second.
- Instantaneous acceleration. Shrink Δt towards zero: a = dv/dt. On a velocity–time graph this is the slope of the tangent at that instant.
- Constant acceleration. Then v = v₀ + at. Starting at 5 m/s with a = 2 m/s², after 3 s: v = 5 + 2×3 = 11 m/s.
The four straight-line graphs in your NCERT book
With constant acceleration a velocity–time graph is a straight line, and its slope is the acceleration. Pick a case:
Common mistakes (examiners love these)
- "Zero velocity means zero acceleration." No: at the top of a throw, v = 0 but a = g.
- "Negative acceleration means slowing down." Not always. In case c the object speeds up with negative acceleration. The rule: if speed increases, acceleration points along the velocity; if speed decreases, it points against it.
- Mixing up the graphs. On a position–time graph the slope is velocity. On a velocity–time graph the slope is acceleration and the area is displacement.
Bonus: the area under a v–t graph is displacement
Constant velocity u for time T gives a flat line, and the area under it is a rectangle, u × T. So 5 m/s for 4 s means 20 m. Why can an area be a distance? Check the dimensions: velocity [L T⁻¹] × time [T] = [L], a length.
| Graph | Slope gives | Area gives |
|---|---|---|
| Position–time (x–t) | Velocity | Not used |
| Velocity–time (v–t) | Acceleration | Displacement |
Check yourself
1. A velocity–time graph is a straight line from 12 m/s at t = 0 to 0 m/s at t = 4 s. What is the acceleration at t = 4 s?
Slope = Δv/Δt = (0 − 12)/4 = −3 m/s². Zero is the velocity at that instant, not the acceleration; +3 misses the sign; −48 multiplies instead of dividing.
2. An object moves in the negative direction and its acceleration is negative. What happens to its speed?
Velocity and acceleration point the same way (both negative), so the object speeds up. This is case c.
3. A cyclist rides at a steady 5 m/s for 4 s. What does the area under her v–t graph give?
Area = 5 m/s × 4 s = 20 m. The area under a velocity–time graph is displacement.
Want a 60-second version? Try the daily quiz Shorts and answer in the comments →
Questions students ask
Is acceleration zero at the highest point of a vertical throw?
No. Velocity is zero for an instant, but acceleration is still g ≈ 9.8 m/s² downward. Otherwise the ball would stay at the top.
What is acceleration, in one line?
The rate of change of velocity with time: a = Δv/Δt on average, or dv/dt at an instant. SI unit m/s².
Does negative acceleration always mean slowing down?
No. It means slowing down only when the velocity is positive. If the velocity is also negative, the object speeds up.
Can an object have zero velocity and non-zero acceleration?
Yes: a ball at the top of a throw, or any object at the instant it turns back while accelerating (case d in NCERT).
Why are real v–t graphs smooth, not sharp-cornered?
Velocity and acceleration can't change abruptly at a single instant in real motion, so real graphs have no sharp kinks. Kinked graphs in the book are simplifications.
One clear physics lesson, every day
School physics explained with animations, plus a quiz Short every day.
Full transcript
Read the whole lesson as text
A ball thrown straight up stops at the very top, for one instant. Its velocity is zero. So is its acceleration zero too? It feels like it should be. By the end, one straight line on a graph will settle that, and a graph like it will tell you how far an object has moved.
Per metre, or per second? Velocity usually changes during motion. How do we describe that change? As the rate of change of velocity with distance, or with time? This was a problem even in Galileo's time. It was first thought to be with distance. But Galileo studied freely falling objects, and objects on an inclined plane. He concluded that for all objects in free fall, the rate of change of velocity with time is constant. The change with distance is not. It decreases as the object falls further. Ignore air resistance, and a falling object gains about 9.8 metres per second, every second. But each second, it also falls further than the second before. So the same gain is spread over more and more metres. This led to acceleration, the rate of change of velocity with time.
Average acceleration. Average acceleration is the change of velocity divided by the time interval, Δv/Δt. Its SI unit is metres per second squared. Say a car is doing 10 m/s at 2 s and 20 m/s at 7 s. The velocity changes by 10 in 5 seconds: average acceleration 2 m/s². On a velocity–time graph, average acceleration is the slope of the straight line joining those two points. Instantaneous acceleration: shrink Δt and take the limit, dv/dt, the slope of the tangent to the velocity–time curve at that instant.
Positive, negative or zero. Velocity has magnitude and direction, so acceleration may come from a change in speed, a change in direction, or both. A position–time graph curves upward for positive acceleration, downward for negative, and is a straight line for zero acceleration. With constant acceleration, v = v₀ + at. Start at 5 m/s with 2 m/s²; after 3 s, 11 m/s.
Four straight lines. Case a: positive direction, positive acceleration, above the axis and rising, speeding up. Case b: positive direction, negative acceleration, above the axis and falling, slowing down. Case c: negative direction, negative acceleration, below the axis and going further down: negative acceleration, yet speeding up. If speed increases, acceleration is along the velocity; if speed decreases, it is opposite. Case d: moves in the positive direction till t₁, then turns back with the same negative acceleration. At t₁ the velocity is zero, but the line does not bend: same slope, same acceleration. That is our ball: at the top its velocity is zero and its acceleration is still g.
Area under the line. The area under a velocity–time curve gives the displacement. Constant velocity u for time T: a rectangle, u × T. Five metres per second for four seconds gives twenty metres. Dimensions: [L T⁻¹] × [T] = [L]. Real graphs are smooth, because velocity and acceleration cannot change abruptly at an instant.
Summary. Acceleration is the rate of change of velocity with time. On a velocity–time graph, slope gives acceleration and area gives displacement.