Equations of uniformly accelerated motion
Three equations from one graph
A car starts with velocity u and has a constant acceleration a. Its v–t graph is a straight line that starts at height u.
The slope of this line is a. So after time t, the velocity has grown by at. That gives our first equation: v = u + at.
Now the displacement x is the area under the line from 0 to t. Split it into two pieces: a rectangle and a triangle on top.
The rectangle is u tall and t wide, so its area is ut. The triangle is t wide and at tall, so its area is ½ × t × at, which is ½at².
Add them up: x = ut + ½at². This is the second equation.
The area is also a trapezium: average of the two sides, (u + v)/2, times t. Swap t for (v − u)/a and tidy up. You get v² = u² + 2ax.
Three equations, one for each job. Use v² = u² + 2ax when time is not given and not asked. Remember: they only work when the acceleration stays constant.
Free fall
Drop a stone and a cricket ball together. Ignoring air resistance, they fall side by side, speeding up at the same rate: g = 9.8 m/s², straight down.
Choose upward as positive. Then the acceleration is a = −g = −9.8 m/s², whether the object is going up, coming down, or at the top.
A ball is thrown straight up at 20 m/s from the top of a 25 m tall building. How high does it rise?
At the top, v = 0. Using v² = u² + 2ay: 0 = 20² − 2 × 9.8 × y, so y = 400 / 19.6, about 20.4 m above the roof.
When does it hit the ground? Take the ground as y = 0. The ball starts at y₀ = 25 m, so y = 25 + 20t − 4.9t². Set y = 0 and solve.
The quadratic gives t ≈ 5.1 s. The other root is negative, so we throw it away. Notice that we never had to split the trip into up and down.
Stopping distance
A car moves at speed u. The driver brakes hard and the car slows down with a constant deceleration of size a. How far does it go before it stops?
At the end v = 0, and the acceleration is −a. Put these into v² = u² + 2ax: 0 = u² − 2a·d. So the stopping distance is d = u² / 2a.
At 36 km/h, that is 10 m/s, with a = 5 m/s², the car stops in 100 / 10 = 10 m.
Now double the speed to 72 km/h, or 20 m/s. The stopping distance becomes 400 / 10 = 40 m. Four times as far, not twice.
And the driver does not brake at once. During the reaction time, maybe 0.7 s, the car keeps going at full speed: another 14 m at 20 m/s.
This is why speed limits near schools are so low. Stopping distance grows with the square of the speed.
Try it: braking car
- The brakes option keeps the car stopped once v = 0, as a real car would.
- The shaded triangle under the v–t graph is the stopping distance: ½ × u × (stopping time).
- Untick the brakes option to see uniform acceleration carry on, like a ball thrown up and falling back.
- Check v = u + at: read v at any time and compare.
Quick check
A car travelling at 10 m/s needs 10 m to stop with its brakes fully on. With the same braking, how far does it need if it is travelling at 20 m/s?
- 10 m
- 20 m
- 30 m
- 40 m
Answer
D. Stopping distance d = u² / 2a. Doubling u makes u² four times bigger, so d = 4 × 10 = 40 m. (Here a = 100 / 20 = 5 m/s².)
Free fall
A stone is dropped from rest off a bridge and hits the water 2 s later. How high is the bridge above the water? (g = 9.8 m/s², ignore air resistance)
- 9.8 m
- 19.6 m
- 39.2 m
- 4.9 m
Answer
B. Dropped means u = 0. Distance fallen = ½gt² = ½ × 9.8 × 2² = 19.6 m. It hits the water at v = gt = 19.6 m/s.
Summary
- Velocity–time:
v = u + at - Position–time:
x = ut + ½at² - Velocity–position:
v² = u² + 2ax - Free fall (upward positive):
a = −g = −9.8 m/s² - Stopping distance:
d = u² / 2a
- The three equations hold only for constant acceleration. x here is the displacement from the starting point.
- Pick a positive direction first and stick to it. Signs of u, v, a and x all follow from that choice.
- In free fall the acceleration is g downwards all the time: going up, at the top, and coming down.
- Stopping distance grows as u²: double the speed, four times the distance.