A car is moving with a velocity of 30 ms -1. On applying the brakes, the velocity decreases to 15 ms -1 in 2 s. The acceleration of the car is

Question

A car is moving with a velocity of 30 ms -1. On applying the brakes, the velocity decreases to 15 ms -1 in 2 s. The acceleration of the car is (A) +7.5 ms -2 (B) -7.7 ms -2 (C) -7.5 ms -2 (D) +15 ms -2

Tardigrade Admin · Accepted Answer

Given, v=15 ms -1, v0=30 ms -1 and t=2 s Using relation, v=v0+a t Acceleration of the car, a =(v-v0/t)=((15-30) ms -1/2 s ) =-(15/2) ms -2=-7.5 ms -2

Q. A car is moving with a velocity of $30 m s^{- 1}$ . On applying the brakes, the velocity decreases to $15 m s^{- 1}$ in $2 s$ . The acceleration of the car is

$+ 7.5 m s^{- 2}$

$- 7.7 m s^{- 2}$

$- 7.5 m s^{- 2}$

$+ 15 m s^{- 2}$

Given, $v = 15 m s^{- 1}, v_{0} = 30 m s^{- 1}$ and $t = 2 s$
Using relation, $v = v_{0} + a t$
Acceleration of the car,
$a = \frac{v - v _{0}}{t} = \frac{( 15 - 30 ) m s ^{- 1}}{2 s}$
$= - \frac{15}{2} m s^{- 2} = - 7.5 m s^{- 2}$

Q. A car is moving with a velocity of 30ms−1. On applying the brakes, the velocity decreases to 15ms−1 in 2s. The acceleration of the car is

+7.5ms−2

−7.7ms−2

−7.5ms−2

+15ms−2