Question

A car is moving with uniform speed of 27m/s. it is brought to rest by applying brakes to cause a retardation of 0.9m/s^2 . calculate 1 - the distance covered by the car before it comes to rest . 2- the time for which the car moves before it comes to rest

Answer 1

Explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• Initial velocity = 27m/s

• Retardation = 0.9m/s²

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• The distance covered by the car

• Time taken by the car

${\green{\underline{\underline{\bold{Solution:-}}}}}$

Intial velocity (u) = 27m/s

Acceleration = -0.9m/s² (As Retardation = 0.9m/s²)

Final velocity (v) = 0m/s

Let the distance covered be ' x '

According to the third equation of motion:-

$\boxed {\sf \pink{ \rightarrow {v}^{2} - {u}^{2} = 2as}}$

Substituting the values:-

${\sf { \rightarrow {0}^{2} - {27}^{2} = 2( - 0.9)s}}$

${\sf { \rightarrow - {27}^{2} = - 1.8s}}$

${\sf { \rightarrow - 729 = - 1.8s}}$

${\sf { \rightarrow \not - 729 = \not - 1.8s}}$

${\sf { \rightarrow \dfrac{729}{1.8} = s}}$

${\sf { \rightarrow 405 = s}}$

Distance = 405m

According to the first equation of motion:-

$\boxed {\sf \purple{ \rightarrow {v} = {u} + at}}$

${\sf { \rightarrow {0} = 27 - 0.9t}}$

${\sf { \rightarrow - 27 = - 0.9t}}$

${\sf { \rightarrow \dfrac{ - 27}{ - 0.9} = t}}$

${\sf { \rightarrow 30 = t}}$

Time = 30sec