Question

A car moving along a straight road with speed of 120 km/hr is brought to rest by applying brakes. The car covers a distance of 100 m before it stops . Calculate the 1.average retardation of the car .... 2.time taken by the car come to rest

Answer 1

Answer:

1. $5.55m | s {}^{2}$
2. t = 6sec

Explanation:

From the question,

u = 120km/hr = 120×5/18 = 33.33m/s

v = 0 m/s { as the car comes to rest }

s = 100m

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We know that;

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

On rearranging,

$a = v {}^{2} - u {}^{2} \div 2 s$

On substituting,

a = 0 - 33.33^2 / 200

= - 1111.11 / 200 {approx value}

= 5.55m/s^2

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Also, we know that,

$v = u + at$

On substituting,

t = v - u /a

= - 33.33 / - 5.55

= 6.005 = 6sec

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Hope it helps....!

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Answer 2

Answer:

Explanation:

Hope it helps