The brakes applied to a car produce an acceleration of $6 {m/s}^{2}$ in the opposite direction to the motion. If the car takes $2 s$ to stop after the application of brakes, calculate the distance it travels during this time.

15 m

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16 m

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20 m

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12 m

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Solution

The correct option is D $12 m$
Given, $a = - 6 {m/s}^{2}, t = 2 s$ and $v = 0 m/s$
From first equation of motion, $v = u + a t$
$0 = u + (- 6) \times 2$
$u = 12 m/s$
Distance being travelled, $s = u t + \frac{1}{2} a t^{2}$
$= 12 \times 2 + \frac{1}{2} \times (- 6) \times (2)^{2}$
$= 24 - 12 = 12 m$

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Similar questions

The brakes applied by a car produces an acceleration of $6 m / s^{2}$ in opposite direction of motion. If the car takes 2 sec to stop after the application of brakes. Calculate distance during this time.

Q. The brakes applied to a car produce an acceleration of

6 {m/s}^{2}

in the opposite direction to the motion. If the car takes

2 s

to stop after the application of brakes, calculate the distance it travels during this time.

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The brakes applied to a car produce an acceleration of 6 m/s2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.

The correct option is D 12 mGiven, a=−6 m/s2,t=2 s and v=0 m/s From first equation of motion, v=u+at 0=u+(−6)×2 u=12 m/s Distance being travelled, s=ut+12at2 =12×2+12×(−6)×(2)2 =24−12=12 m

The brakes applied to a car produce an acceleration of $6 {m/s}^{2}$ in the opposite direction to the motion. If the car takes $2 s$ to stop after the application of brakes, calculate the distance it travels during this time.