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Question

If the speed of the particle is v=8t33t2+2 m/s, the distance of the particle as a function of time t and the average speed in 2 sec respectively are

A
2t4t3+2t m,14 m/s
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B
2t4t3+t m,14 m/s
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C
2t4t3+2t m,28 m/s
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D
t4t3+2t m,14 m/s
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Solution

The correct option is A 2t4t3+2t m,14 m/s
Given, the speed of the particle is v(t)=8t33t2+2
So, the distance as a function of time is given as
v(t)dt
S=(8t33t2+2)dt=2t4t3+2t m
Thus, the distance travelled in 2 sec is S=2×2423+2×2=28 m
Hence, average speed of the particle is
Total distance travelled (S)Total time taken (t)=282=14 m/s

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