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When a vehicle suddenly starts or stops, you feel a jolt — the feeling of velocity changing. How do we measure how quickly velocity changes?
- The average acceleration over a time interval is the change in velocity divided by the time interval: $$a = \dfrac{\text{change in velocity}}{\text{time interval}} = \dfrac{v - u}{t_2 - t_1}$$ where the velocity changes from \(u\) (at \(t_1\)) to \(v\) (at \(t_2\)).
- SI unit: m s⁻² (m/s²). It has magnitude and direction.
- If the magnitude of velocity is increasing , acceleration is in the direction of velocity; if decreasing , it is opposite to velocity.
- Average acceleration — the change in the velocity of an object over a time interval divided by that time interval; its SI unit is m s⁻².
- The reading of a vehicle’s speedometer is nearly (but not exactly) the magnitude of the velocity at an instant.
- The direction of the tyres gives the direction of velocity at that instant.
In this Activity, we will use the 0-to-100 km/h time of different cars (looked up online) to calculate and compare their average accelerations.
- The acceleration of a car is often quoted as the time to go from 0 to 100 km h⁻¹. Note this time for a few cars, then compute \(a\). (Note: \(100\ \text{km h}^{-1} \approx 27.8\ \text{m s}^{-1}\).)
| Car type (illustrative) | Time from 0 to 100 km h⁻¹ | Magnitude of average acceleration (m s⁻²) |
|---|---|---|
| Sports car | 5 s | ≈ 5.6 |
| Sedan | 8 s | ≈ 3.5 |
| Hatchback | 12 s | ≈ 2.3 |
A bus at 36 km h⁻¹ speeds up to 54 km h⁻¹ in 10 s, then later brakes from 54 km h⁻¹ to rest in 5 s. Find the average acceleration in each interval.
(i) Accelerator: \(u = 36\ \text{km h}^{-1} = 10\ \text{m s}^{-1}\), \(v = 54\ \text{km h}^{-1} = 15\ \text{m s}^{-1}\), \(t = 10\) s.
\( a = \dfrac{15 - 10}{10} = \dfrac{5}{10} = 0.5\ \text{m s}^{-2} \) (in the direction of velocity, since speed is increasing).
(ii) Brake: \(u = 15\ \text{m s}^{-1}\), \(v = 0\), \(t = 5\) s.
\( a = \dfrac{0 - 15}{5} = -3\ \text{m s}^{-2} \). The minus sign shows acceleration is opposite to velocity (speed decreasing).
A dropped object has velocity 9.8, 19.6, 29.4, 39.2 m s⁻¹ at t = 1, 2, 3, 4 s. Find its average acceleration in each one-second interval. Is it constant?
\( \dfrac{9.8 - 0}{1} = 9.8,\quad \dfrac{19.6 - 9.8}{1} = 9.8,\quad \dfrac{29.4 - 19.6}{1} = 9.8,\quad \dfrac{39.2 - 29.4}{1} = 9.8\ \text{m s}^{-2} \)
The average acceleration is constant, 9.8 m s⁻² , in the direction of motion. This is the acceleration due to gravity, g .
- An object can move very fast and still have zero acceleration . Acceleration depends on how quickly velocity changes , not on how fast the object is going.
- Average acceleration can come from a change in the magnitude of velocity, its direction, or both. In this chapter we consider only cases where acceleration is constant .
- Just like velocity at an instant, the acceleration at a particular instant is called the instantaneous acceleration .
- As the time interval around an instant becomes very small, the average acceleration approaches a fixed value — the instantaneous acceleration. You will learn more about it in higher grades.
