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For constant acceleration, can we write equations that let us predict an object’s velocity and position at any future time?
- Relating displacement \(s\), time \(t\), initial velocity \(u\), final velocity \(v\) and constant acceleration \(a\):
- $$v = u + at \quad \text{(4.4a)}$$
- $$s = ut + \tfrac{1}{2}at^2 \quad \text{(4.4b)}$$
- $$v^2 = u^2 + 2as \quad \text{(4.4c)}$$
- These let us predict the position or velocity of an object at a future time, only when the acceleration is constant .
- From the definition of acceleration \(a = \dfrac{v-u}{t}\), rearranging gives \(at = v - u\), so $$v = u + at \quad \text{(4.4a)}$$
- Displacement = area under the velocity-time line = rectangle + triangle: \( s = u\,t + \tfrac{1}{2}\,t\,(v-u) \). Substituting \(v - u = at\): $$s = ut + \tfrac{1}{2}at^2 \quad \text{(4.4b)}$$
- Eliminating \(t\) between (4.4a) and (4.4b) gives \(v^2 = u^2 + 2as\) (4.4c). The remaining two equations are derived in The Journey Beyond.
- Kinematic equations — the set v = u + at, s = ut + ½at², v² = u² + 2as that relate displacement, time, initial and final velocity and acceleration for motion in a straight line with constant acceleration .
- The equations are valid only for constant acceleration . For one-direction motion, distance = magnitude of displacement and speed = magnitude of velocity.
- For motion in both directions, the signs of \(u, v, a, s\) carry the direction.
Brakes give an acceleration of −4 m s⁻². How far does the car travel before stopping if it was moving at (i) 54 km h⁻¹ and (ii) 108 km h⁻¹?
Given \(a = -4\ \text{m s}^{-2}\), \(v = 0\). Using \(v^2 = u^2 + 2as\): \(0 = u^2 - 8s \Rightarrow s = \dfrac{u^2}{8}\).
(i) \(u = 54\ \text{km h}^{-1} = 15\ \text{m s}^{-1}\): \( s = \dfrac{15^2}{8} = \dfrac{225}{8} \approx 28.1\ \text{m} \).
(ii) \(u = 108\ \text{km h}^{-1} = 30\ \text{m s}^{-1}\): \( s = \dfrac{30^2}{8} = \dfrac{900}{8} = 112.5\ \text{m} \).
Doubling the speed makes the stopping distance about four times larger.
- A braking vehicle still travels some distance before stopping. That distance depends on the speed , the road surface (wet/dry), the braking capacity , and the driver’s reaction time .
- This is why we keep a safe distance from the vehicle ahead, adjusted to our speed. New vehicle-to-vehicle (V2V) technology can warn drivers of possible collisions.
NCERT Question 4 — A car starts from rest
A car reaches 24 m s⁻¹ in 6 s from rest. Find the average acceleration and the distance travelled in these 6 s.
View the answer →NCERT Question 5 — A motorbike moving with initial
A motorbike with u = 28 m s⁻¹ and constant acceleration stops after 98 m. Find the acceleration and the time to stop.
View the answer →NCERT Question 8 — A truck driver driving at
A truck slows from 54 km h⁻¹ to 36 km h⁻¹ in 36 s (constant acceleration). Find the distance travelled.
View the answer →NCERT Question 9 — A car starts from rest
A car speeds up to 20 m s⁻¹ in 5 s, runs at 20 m s⁻¹ for 10 s, then brakes to stop in 6 s. Find the total distance.
View the answer →NCERT Question 10 — A bus is travelling at
A bus at 36 km h⁻¹ sees an obstacle 30 m ahead; reaction time 0.5 s, then a = 2.5 m s⁻² deceleration. Will it stop in time?
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