Answer to Question #12828 in Mechanics | Relativity for akhil madhu
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.
1
2012-08-10T09:43:14-0400
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.
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