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The position x of a particle varies with time according to the relation x=t3+3t2+2t. Find velocity and acceleration as functions of time.

Answer
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Hint: First derivative of function ‘x’ with respect to time gives velocity and double derivative of ‘x’ gives acceleration.

Complete step-by-step answer:
The position of the particle is given by variable x, and it varies according to time.
Given the relationx=t3+3t2+2t(1)
To find the velocity, which is the rate of change of displacement.
The first derivative of Eqn(1) gives us the velocity and the second derivation will give the acceleration.
Velocity =dxdt
v=ddt(x)=ddt(t3+3t2+2t)v=3t2+2(3t)+2v=3t2+6t+2
The unit of velocity is meter per second (m/sec).
v=(3t2+6t+2)m/sec.
To find acceleration, which is the rate of change of velocity.
Acceleration, a=dvdt
a=ddt(v)=ddt(3t2+6t+2)a=2×(3t)+6=6t+6
The unit of acceleration is meter per second square (m/sec2).
a=(6t+6)m/sec2
Velocity of the function, v=(3t2+6t+2)m/sec.
 Acceleration of the function, a=(6t+6)m/sec2

Note: We know velocity=Displacementtimeand acceleration=velocitytime, here the velocity is taken as the rate of change of displacement w.r.t the time, so differentiation (dxdt)is done.
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