Kinematics: Numerical Problems Set (1)

easy • 3 min read
Tags Kinematics

Question

A car starts from rest and accelerates uniformly at 4 m/s24 \text{ m/s}^2 for 55 seconds. Find the final velocity and the distance covered. Then, the driver applies brakes that decelerate the car uniformly to rest over the next 1010 seconds. Find the distance covered during braking and the total distance.

Solution — Step by Step

We use v=u+atv = u + at with u=0u = 0, a=4 m/s2a = 4 \text{ m/s}^2, t=5 st = 5 \text{ s}.

v=0+4×5=20 m/sv = 0 + 4 \times 5 = 20 \text{ m/s}

Using s1=ut+12at2s_1 = ut + \tfrac{1}{2}at^2:

s1=0+12(4)(5)2=50 ms_1 = 0 + \tfrac{1}{2}(4)(5)^2 = 50 \text{ m}

Now u=20 m/su = 20 \text{ m/s}, v=0v = 0, t=10 st = 10 \text{ s}. From v=u+atv = u + at:

0=20+a(10)  ⟹  a=−2 m/s20 = 20 + a(10) \implies a = -2 \text{ m/s}^2

The negative sign confirms deceleration.

Use v2=u2+2as2v^2 = u^2 + 2as_2:

0=(20)2+2(−2)s2  ⟹  s2=100 m0 = (20)^2 + 2(-2)s_2 \implies s_2 = 100 \text{ m} stotal=s1+s2=50+100=150 ms_{\text{total}} = s_1 + s_2 = 50 + 100 = 150 \text{ m}

Final answer: Final velocity after acceleration =20= 20 m/s, total distance =150 m= \mathbf{150 \text{ m}}.

Why This Works

We split the motion into two phases because acceleration changes between them. The three equations of motion only work when acceleration is constant — applying them blindly to the whole journey gives wrong answers.

Once the phases are separated, each one is a straightforward substitution. The final velocity of phase 1 becomes the initial velocity of phase 2, which is the bridge between them.

Alternative Method

We could solve phase 2 using the average velocity shortcut. Since deceleration is uniform, average speed during braking =(20+0)/2=10 m/s= (20 + 0)/2 = 10 \text{ m/s}. Distance =10×10=100 m= 10 \times 10 = 100 \text{ m}. Same answer, faster.

This shortcut works only when acceleration is constant — which is exactly the case here.

Common Mistake

Students often try to use a single equation for the entire motion, treating “acceleration” as a single value. The car accelerates, then decelerates — these are two separate constant-acceleration phases. Mixing them gives nonsense. Always sketch a v-t graph first; the kink shows where you must break the problem.

In CBSE board answers, write the sign of acceleration explicitly. A negative aa during braking earns the conceptual mark even if the arithmetic slips later.

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