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Question

A car moving along a straight highway with a speed of 126km/hr is brought to a stop within a distance of 200m . What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?


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Solution

Step 1. Given data:

Initialvelocity(u)=126Kmh-1Initialvelocity(u)=1263600×1000Initialvelocity(u)=35ms-1

Distance(s)=200m

Finalvelocity(v)=0, since the car will stop after some time

Step 2. Formula used:

3rd Equation of Motion

v2=u2+2as

Where v is the final velocity,

u is initial velocity,

a is acceleration and

s is distance

First Equation of Motion

v=u+at

Where t is the time taken

Step 3. Calculating retardation of the car,
v2=u2+2as

a=-u22s

a=-u22s

=-3.06ms2

The negative sign indicates that the car is retarding.

Retardation is nothing but negative acceleration.

Step 4. Calculating time taken,

v=u+at

t=-ua

=-35-3.06

t=11.43s

Hence, the retardation of car is -3.06ms-2 and time taken by the car to stop is t=11.43s


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