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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
None of 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 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 = 76 m/s and 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 sec = s10s8

= 28.320=8.3 m


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