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Question
A body starts from the rest and accelerates at constant rate of 2 m/s2and then retards at a constant rate of 5 m/s2for some time to come to the rest. Calculate the total distance covered by it if the body remains in motion for 6 seconds.
A body starts from the rest and accelerates at constant rate of 2 m/s2and then retards at a constant rate of 5 m/s2for some time to come to the rest. Calculate the total distance covered by it if the body remains in motion for 6 seconds.
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Solution
Let the body accelerate for a time of t seconds.
Therefore, the body decelerated for a time of 6 - t seconds.
Velocity at the end of t seconds = 2t
2t = 5(6- t)
Therefore, t = 30/7 seconds
Maximum velocity = 2t = 60/7 m/s
Total distance traveled (area of the velocity time graph, in this case a triangle)= (1/2)(6)(2t) = 180/7 m
Let the body accelerate for a time of t seconds.
Therefore, the body decelerated for a time of 6 - t seconds.
Velocity at the end of t seconds = 2t
2t = 5(6- t)
Therefore, t = 30/7 seconds
Maximum velocity = 2t = 60/7 m/s
Total distance traveled (area of the velocity time graph, in this case a triangle)= (1/2)(6)(2t) = 180/7 m
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