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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 displacement 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 s of motion s5=10 m , so by using s=ut+12at2
10=5u+12a(5)22u+5a=4(i)
For first time 8 s of motion s8=20 m
20=8u+12a(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+12a(10)2
By substituting the value of u and a we will get s10=28.3 m
So the distance in last 2 s =s10s8=28.320=8.3 m

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