A truck is moving with the speed of 72 km/h. Suddenly, driver
sees an obstacle and he applies the brake to stop the truck in
5 sec. Find the retardation produced by brakes and the
distance travelled by the truck before it stops.
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Answer:
Explanation:
Given,
Speed=20m/s,d=50m,retardation=5m/s 2
Let the ppossible reaction time be t s
So the car moves 20tm And applied breaks with retardation 5m/s2
So, the total distance under retardation is ≤50m
Thus the distance under the application of retardation =
(20)2-(0)2/2*5=s ′ ⇒s ′=40m
Therefore, 20t s+40≤=50⇒20t s
≤50−40⇒20t s
=10⇒t s≤ 1/2≤0.5s
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