Question

If the distance ‘s’ travelled by a particle in time t is given by $s = a\sin t + b\cos 2t$, then the acceleration at $t = 0$ is:
(A) \[a\]
(B) \[ - a\]
(C) \[4b\]
(D) \[ - 4b\]

Answer 1

Hint: Given problem tests the concepts of derivatives and their applications. These type questions can be easily solved if we keep in mind the concepts of derivatives as rate measure. In the problem, we are given the expression for distance travelled by particles as $s = a\sin t + b\cos 2t$. We first find the expression for acceleration by differentiating the expression of distance s two times with respect to time t and then put the value of t as zero to find acceleration at $t = 0$.

Complete step-by-step answer:
So, we have the distance travelled by a particle as $s = a\sin t + b\cos 2t$.
We differentiate the expression with respect to time t. So, we get the speed of particle as,
\[\dfrac{{ds}}{{dt}} = \dfrac{d}{{dt}}\left( {a\sin t + b\cos 2t} \right)\]
\[ \Rightarrow v = \dfrac{d}{{dt}}\left( {a\sin t} \right) + \dfrac{d}{{dt}}\left( {b\cos 2t} \right)\]
Taking constant terms out of differentiation, we get,
\[ \Rightarrow v = a\dfrac{d}{{dt}}\left( {\sin t} \right) + b\dfrac{d}{{dt}}\left( {\cos 2t} \right)\]
We know that the derivative of $\cos x$ is $ - \sin x$ and the derivative of $\sin x$ is \[\cos x\]. Using the chain rule of differentiation $\dfrac{{d\left( {fog\left( x \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$, we get,
\[ \Rightarrow v = a\cos t + b\left( { - \sin 2t} \right) \times 2\]
Simplifying the expression,
\[ \Rightarrow v = a\cos t - 2b\sin 2t\]
Now, differentiating the expression for speed of particle with respect to time t to get acceleration.
$\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}\left( {a\cos t - 2b\sin 2t} \right)$
\[ \Rightarrow A = \dfrac{d}{{dt}}\left( {a\cos t} \right) - \dfrac{d}{{dt}}\left( {2b\sin 2t} \right)\]
Taking constants out of differentiation, we get,
\[ \Rightarrow A = a\dfrac{d}{{dt}}\left( {\cos t} \right) - 2b\dfrac{d}{{dt}}\left( {\sin 2t} \right)\]
Using chain rule of differentiation and substituting derivatives of functions, we get,
\[ \Rightarrow A = a\left( { - \sin t} \right) - 2b\left( {\cos 2t} \right) \times \dfrac{{d\left( {2t} \right)}}{{dt}}\]
\[ \Rightarrow A = - a\sin t - 4b\left( {\cos 2t} \right)\]
Now, we have got the expression for acceleration of the particle. So, we substitute the value of t as zero to find the acceleration at time $t = 0$.
\[ \Rightarrow A\left( {t = 0} \right) = - a\sin 0 - 4b\cos \left( {2 \times 0} \right)\]
Substituting value of $\sin 0$ and $\cos 0$, we get,
\[ \Rightarrow A\left( {t = 0} \right) = - a \times 0 - 4b\cos 0\]
\[ \Rightarrow A\left( {t = 0} \right) = - 4b\left( 1 \right)\]
\[ \Rightarrow A\left( {t = 0} \right) = - 4b\]
So, the value of acceleration at time $t = 0$ is \[ - 4b\].
Therefore, option (D) is the correct answer.
So, the correct answer is “Option D”.

Note: We must know that the acceleration of a particle can be calculated by differentiating the expression of distance twice with respect to time t. We should remember the chain rule and power rule of differentiation. Derivatives of some basic trigonometric functions must be remembered in order to get to the final answer.