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Q.

The position of a particle moving in a straight line is given by x=3t3−18t2+36t. Here x is in m, and ‘t’ in second. Then

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a

Direction of velocity and acceleration both change at t=2 sec

b

The distance travelled by particle is equal to magnitude of displacement for t = 0 to t = 5

c

The speed of particle is decreasing in t= 0 to t = 2sec then it is increasing for t>2

d

The magnitudes of velocity and acceleration are equal at t = 0

answer is B.

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Detailed Solution

x=3t3−18t2+36t v=dxdt=9t−22⇒a=18t−2
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