SOLUTION: A vehicle moves at an initial speed of 20m/sec with a constant acceleration of 2m/sec square, for 5 sec before brakes are applied. if tha car comes to rest under constant decelerat
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Question 1101227: A vehicle moves at an initial speed of 20m/sec with a constant acceleration of 2m/sec square, for 5 sec before brakes are applied. if tha car comes to rest under constant deceleration in 4 sec, determine the total distance travelled during the 9 seconds? Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! Using the formula v2 = v1 + at, the speed after the 5s acceleration is:
v2 = 20 + 2*5 = 30 m/s
Using the formula d = v1*t + 1/2a*t^2, the distance covered during acceleration is:
d1 = 20*5 + 1*5^2 = 125 m
The rate of deceleration is
a = (v2-v1)/t = (0-30)/4 = -7.5 m/s^2
The distance during 4s deceleration is:
d2 = 30*4 - (7.5/2)*4^2 = 60 m
Thus the total distance traveled is 125 + 60 = 185 m
