You are here
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 (4.4a)
- s = ut + 1 2 at 2 (4.4b)
- v 2 = u 2 + 2as (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 = v-u t , rearranging gives at = v - u , so v = u + at (4.4a)
- Displacement = area under the velocity-time line = rectangle + triangle: s = u t + 1 2 t (v-u) . Substituting v - u = at : s = ut + 1 2 at 2 (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 m s -2 , v = 0 . Using v 2 = u 2 + 2as : 0 = u 2 - 8s ⇒ s = u 2 8 .
(i) u = 54 km h -1 = 15 m s -1 : s = 15 2 8 = 225 8 ≈ 28.1 m .
(ii) u = 108 km h -1 = 30 m s -1 : s = 30 2 8 = 900 8 = 112.5 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 starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
NCERT Question 5 — A motorbike moving with initial
A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
NCERT Question 8 — A truck driver driving at
A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ (Fig. 4.29) for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
NCERT Question 9 — A car starts from rest
A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
NCERT Question 10 — A bus is travelling at
A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s⁻². Will the bus be able to stop before reaching the obstacle?
-
Write the three kinematic equations for constant acceleration.
Show Answer
v = u + at · s = ut + ½at² · v² = u² + 2as -
A body starts from rest with a = 2 m s⁻². What is its velocity after 5 s?
Show Answer
v = u + at = 0 + (2)(5) = 10 m s⁻¹. -
Which equation would you use if time is not given?
Show Answer
v² = u² + 2as — it links velocity, acceleration and displacement without time. -
When are these three equations valid?
Show Answer
Only for motion in a straight line with a constant (uniform) acceleration.
