Physics, asked by navzjaglan, 1 year ago

A body starts from rest moves with uniform acceleration of 10m/s2 for 10second then it deaccelerates by 5m/2 and comes to rest .the distance covered by the body is

Answers

Answered by Ashugupta942
0
v=u+at
v=10×10=100m/s
v=u+at
0=100+5×t
t=20s
total distance=1600
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