    Question

# A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec ​

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Solution

## The correct option is D Let initial (t=0) velocity of particle = u For first 5 sec motion s5=10 metre s=ut+12at2⇒10=5u+12a(5)2 2u+5a=4.....(i) For first 8 sec of motion s8 = 20 metre 20=8u+12a(8)2⇒2u+8a=5.....(ii) By solving u=76ms and a=13ms2 Now distance travelled by particle in Total 10 sec. s10=u×10+12a(10)2 By substituting the value of u and a we will get s10=28.3m so the distance in last 2 sec = s10−s8 =28.3−20=8.3m  Suggest Corrections  0      Similar questions  Related Videos   Motion Under Constant Acceleration
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