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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
10.3 m
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C
9.3 m
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D
None of the above
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Solution

The correct option is A 8.3 m
Step: 1 Find the relation of initial velocity and acceleration of the particle in first 5 sec.
Formula Used: S=ut+12at2
Given that particle travels 10 m in first 5 sec
Let initial (t=0) velocity of particle u
For first 5 sec motion s5=10 metre
s5=ut+12at2
10=5u+12a(5)2
2u+5a=4 (i)

Step: 2 Find the relation of initial velocity and acceleration of the particle in first 8 sec.
Formula Used: S=ut+12at2
For first 8 sec of motion, s8=20 meter.
20=8u+12a(8)2
2u+8a=5 (ii))

From equation (𝑖) and (𝑖𝑖), we get
u=76 m/s and a=13 m/s2

Step: 3 Find the distance travelled by the particle in 10 sec.
Now distance travelled by particle in total 10 sec
s10=u×10+12a(10)2
s10=76×10+12(13)(10)2 s=28.3 m

Step: 4 Find the distance travelled by the particle in last 2 sec.
The distance in last 2 sec
=s10s8
=28.320
=8.3 m

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