CBSE Class 11-science Answered
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Asked by rajeshabirami27 | 07 May, 2024, 16:31: 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, 18:33: PM
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