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Asked by rajeshabirami27 | 07 May, 2024, 04:31: PM

Let the speed u of at first point is u m/s .

After 6 s , it reaches the point with speed v = 45 m/s with constant accelereation a.

Then we have ,  v = u + at

45 = u + 6 a ...................(1)

After 6 s , it travelled a distance S = 180 m/s with constant acceleration a to reach the second point .

Then we have  ,

S = u t + (1/2 ) a t2

180 = 6 u + (1/2) a ( 36 )

30 = u + 3 a ................(2)

By solving eqn.(1) and (2) , we get

u = 15 m/s  and a = 5 m/s2

Answered by Thiyagarajan K | 07 May, 2024, 06:33: PM

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