A particle moves in a straight line according to the relation x=t3-4t2+3t. Find the acceleration of the particle at displacement equal to zero
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there are 3 instants when displacement of the body is 0 i.e. at t=0, t=1, t=3. By double differentiating the given eqn. we get the eqn. for acceleration i.e. a=6t-8----->(1). Substituting the values of t in (1), we get a=-8 for t=0, -2 for t=1, 10 for t=3.
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Answer:
a=-8 for t=0,
a=-2 for t=1,
a= 10 for t=3.
Explanation:
Given: x=t3-4t2+3t
Acceleration is the rate at which velocity changes with time.
differentiating x with respect to t,
dx/dt= 3t2-8t
again differentiating dx/dt with respect to t,
d2x/dt2 = 6t-8
Now we will put t=0,1,2
putting t=0 , a= -8
t=1, a= -2
t=2, a = 3
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