# Answer to Question #12828 in Mechanics | Relativity for akhil madhu

Question #12828

a motor car slows down from 72km/h to 332km/h over a distance of 25m.if the breaks are applied with the same force.caculate:-

i) total time in which it comes to rest.

ii)distance travelled by it.

i) total time in which it comes to rest.

ii)distance travelled by it.

Expert's answer

i) total time in which it comes to rest.

Let's find the deceleration of a car. The deceleration time is

t = 0.025km / ((72km/h-32km/h)/2) = 0.00125h.

Deceleration of a car is

a = (72km/h-32km/h) / 0.00125h = 32000km/h².

So, it will come to rest in

T = 72km/h / 32000km/h² = 0.00225h.

ii)distance traveled by it.

L = 0.00225h * (72km/h/2) = 0.081km.

Let's find the deceleration of a car. The deceleration time is

t = 0.025km / ((72km/h-32km/h)/2) = 0.00125h.

Deceleration of a car is

a = (72km/h-32km/h) / 0.00125h = 32000km/h².

So, it will come to rest in

T = 72km/h / 32000km/h² = 0.00225h.

ii)distance traveled by it.

L = 0.00225h * (72km/h/2) = 0.081km.

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