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Question

A particle travels 10 m in first 5 sec and 10 m in next 3 sec. 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
28.3 m
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Solution

The correct option is A 8.3 m
Let initial velocity of particle be =u
For first 5 sec motion, s5=10 metre
As we know that
s=ut+12at210=5u+12a(5)2
2u+5a=4 ... (i)
For first 8 sec of motion s8=20 metre
Similarly,
20=8u+12a(8)22u+8a=5 ... (ii)
By solving the above two equations, we get
u=76 m/s and a=13 m/s2
Now distance travelled by particle in total 10 sec is given by
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 sec=s10s8
=28.320=8.3 m

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