If the speed of the particle is v=8t3−3t2+2m/s, the distance of the particle as a function of time t is
A
2t4−2t3+2tm
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B
t4−t3+2tm
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C
2t4−t3+2tm
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D
2t4−t3+2m
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Solution
The correct option is C2t4−t3+2tm Given, the speed of the particle is v(t)=8t3−3t2+2 So, the distance as a function of time is given as ∫v(t)dt ⇒S=∫(8t3−3t2+2)dt=2t4−t3+2tm