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Question

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

A
8.3 m
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B
9.3 m
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C
10.3 m
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D
None of the above
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Solution

The correct option is A 8.3 m
Let initial (t=0) velocity of particle = u
For first 5 sec of motion s5=10 m, using s=ut+12at2
10=5×u+12×a×(5)22u+5a=4(i)
For first time 8 sec of motion s8=10+10=20 m
20=8×u+12×a×(8)22u+8a=5(ii)
By solving (i) and (ii) u=76 m/s a=13 m/s2
Now distance travelled by particle in total 10 sec, s10=u×10+12×a×(10)2
By substituting the value of u and a we will get s10=28.3 m
So the distance in last 2 sec =s10s8=28.320=8.3 m

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