Question

The velocity of a particle is given by \[V=12+3\left( t+7{{t}^{2}} \right)\]. What is the acceleration of the particle?
A. 3+21t
B. 3+42t
C. 42t
D. 4t

Answer 1

Hint: Acceleration is the rate of change of velocity. Differentiate the given expression with respect to t. Simplify the expression to obtain the acceleration of the particle.

Complete step-by-step answer:
Here, we are given the velocity of a particle with time t.
\[V=12+3\left( t+7{{t}^{2}} \right)-(1)\]
We know acceleration is the rate of change of velocity. Like velocity acceleration is a vector and has both magnitude and direction. This is something that we have learned in physics.

For example, a car in straight line motion is said to have forward (positive), acceleration if it is speeding up and backward (negative) acceleration if it is slowing down.
Thus acceleration is the rate of change of velocity.
To find acceleration of the particle we need to differentiate equation (1). Differentiating the equation with respect to time ‘t’, we get the rate of change of velocity, which is acceleration.
Therefore, we need to find, \[\dfrac{dV}{dt}\].
\[\begin{align}
 & \dfrac{dV}{dt}=\dfrac{d}{dt}\left( 12+3\left( t+7{{t}^{2}} \right) \right) \\
 & \dfrac{dV}{dt}=\dfrac{d}{dt}12+3\dfrac{d}{dt}\left( t+7{{t}^{2}} \right) \\
\end{align}\]
\[\begin{align}
 & \dfrac{dV}{dt}=0+3\left( 1+7\left( 2t \right) \right) \\
 & \dfrac{dV}{dt}=3\left[ 1+14t \right] \\
\end{align}\]
\[\begin{align}
 & \dfrac{dV}{dt}=3+14\times 3t \\
 & \dfrac{dV}{dt}=3+42t \\
\end{align}\]
Hence we got acceleration of the particle as \[\left( 3+42t \right)\].
So, option B is the correct answer.
Note: The key to solve these types of problems is to remember the theory of displacement, velocity and acceleration. Velocity of a particle is the rate of change of displacement with respect to time t. Similarly, acceleration of a particle is the rate of change of velocity w.r.t time.