Question

A car is moving at 120 kmh^-1. On seeing a speed breaker, the driver applies the brakes and stops the car in 4 s. What is the value of the deceleration produced in the car is _________ ms^-2

• A. $-\frac{175}{5}$
• B. $+\frac{25}{3}$
• C. $-\frac{3}{25}$
• D. $-\frac{25}{3}$

Answer 1

The correct option is B

$+\frac{25}{3}$

Step 1: Given data

the initial speed of the car U= $120\frac{km}{h}=\frac{120\times 1000}{60\times 60}=\frac{100}{3}\frac{m}{s}$

the final velocity V= 0 ms^-1

time taken t= 4 sec

Step 2: Calculation

From the kinematic equation

$$\begin{gathered} V=U+at \\ 0=\frac{100}{3}+4a \\ -4a=\frac{100}{3} \\ -a=\frac{100}{3\times 4} \\ -a=\frac{25}{3}m{s}^{-2} \end{gathered}$$

If the value of acceleration decreases with an increase in time then such accelerations are called deceleration or retardation.

Deceleration is a value of acceleration that is negative.

So, Deceleration= $\frac{25}{3}m{s}^{-2}$

Hence, deceleration is $\frac{25}{3}\frac{m}{s^2}$and option (b) is correct.