Question

A car starts moving with uniform acceleration from its position of rest and
it moves 100 m in 10 s. On applying brakes, it stops after covering 50 m.
Then magnitude of acceleration in the second part of its motion is _________
m s–2.

• A. 20
• B. 200
• C. 40
• D. 4

Answer 1

The correct option is D

4

Step 1: Calculating acceleration before retardation starts.

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 100=0+\frac{1}{2}a\left(100\right) \\ 2m{s}^{-1}=a \end{gathered}$$, here s is distance, u is initial velocity and t is the time taken

Step 2: Calculating initial velocity when retardation just starts.

$$\begin{gathered} v=u+at \\ v=0+2\left(10\right) \\ v=20m{s}^{-1} \end{gathered}$$

this velocity serves as the initial velocity when deacceleration starts, v=u'

Step 3:

$$\begin{gathered} 2a\text{'}s\text{'}=v{\text{'}}^2-u{\text{'}}^2 \\ 2a\text{'}\left(50\right)=0-u{\text{'}}^2 \\ 100a\text{'}=-400 \\ a\text{'}=-4m{s}^{-1} \end{gathered}$$

the negative sign indicates retardation.

Hence, Option D is the correct answer