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 mStep: 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 =s10−s8 =28.3−20 =8.3 m

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