A car is moving at a velocity of 54Kmh ^ -1. Suddenly driver applies b

1.	A car is moving at a velocity of 54Kmh ^ -1. Suddenly driver applies brakes and produces a deceleration of 6m/s2. Find the distance covered by the car before coming to rest.
Answer» Given:- Initial VELOCITY of the car(u)=54km/h =>54×5/18 =>3×5 =>15m/s Deceleration=6m/s Thus,ACCELERATION(a) is -6m/s² Final velocity(v)=0 (as the car finally STOPS after the application of brakes) To find:- Distance covered by the car be fore COMING to rest(s). Solution:- We wil use here the third equation of motion i.e.---- ${v}^{2} - {u}^{2} = 2as$ $0 - {15}^{2} = 2 \times ( - 6)s$ $- 225 = - 12s$ $s = \frac{ - 225}{ - 12}$ $s = 18.75m$ Thus,the distance covered by the car is 18.75m.

A car is moving at a velocity of 54Kmh ^ -1. Suddenly driver applies brakes and produces a deceleration of 6m/s2. Find the distance covered by the car before coming to rest.

Answer»

Given:-

Initial VELOCITY of the car(u)=54km/h

=>54×5/18

=>3×5

=>15m/s

Deceleration=6m/s

Thus,ACCELERATION(a) is -6m/s²

Final velocity(v)=0 (as the car finally STOPS after the application of brakes)

To find:-

Distance covered by the car be fore COMING to rest(s).

Solution:-

We wil use here the third equation of motion i.e.----

${v}^{2} - {u}^{2} = 2as$

$0 - {15}^{2} = 2 \times ( - 6)s$

$- 225 = - 12s$

$s = \frac{ - 225}{ - 12}$

$s = 18.75m$

Thus,the distance covered by the car is 18.75m.

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