If the distance ‘ $s$ ’ traveled by a particle in time $t$ is $s = a \sin t + b \cos 2 t$ , then the acceleration at $t = 0$ is

$a$

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$- a$

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$4 b$

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$- 4 b$

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Solution

The correct option is D
$- 4 b$

Step 1 Given data:
The distance ‘ $s$ ’ traveled by a particle in time $t$ is $s = a \sin t + b \cos 2 t$
Step 2 Formula used:
Acceleration( $a$ ) is the second derivative of displacement( $s$ ).
That is, $a = \frac{d^{2} s}{{dt}^{2}}$
Step 3 Acceleration at $t = 0$
Acceleration, $a = \frac{d^{2} s}{{dt}^{2}} . . . . (1)$
That is, differentiating the displacement with respect to time
$\frac{ds}{dt} = \frac{d (asint + bcos 2 t)}{dt} \frac{ds}{dt} = acost - 2 b (\sin 2 t)$
Once again, differentiating the displacement with respect to time,
$\frac{d^{2} s}{{dt}^{2}} = \frac{d (acost - 2 b (\sin 2 t))}{dt} \frac{d^{2} s}{{dt}^{2}} = - asint - 4 bcos 2 t . . . (2)$
So, from equation $(1)$ and $(2)$ ,
$a = - asint - 4 bcos 2 t$
Therefore, acceleration at $t = 0$ is,
$a = - asin \times 0 - 4 bcos 2 \times 0 a = a \times 0 - 4 b \times 1 a = 0 - 4 b a = - 4 b$
The distance ‘ $s$ ’ traveled by a particle in time $t$ is $s = a \sin t + b \cos 2 t$ , then the acceleration at $t = 0$ is $- 4 b$ . Hence, option D is correct.

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