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1. ⇒ (MHT CET 2023 10th May Evening Shift )

The displacement ' $S$ ' of a moving particle at a time $t$ is given by $S = 5 + \frac{48}{t} + t^{3}$ . Then its acceleration when the velocity is zero, is

A. 12

B. 20

C. 16

D. 24

Correct Option is (D)

Given, $S = 5 + \frac{48}{t} + t^{3}$

$velocity (V) = \frac{dS}{dt} = 0 - \frac{48}{t^{2}} + 3 t^{2}$

$V = \frac{- 48}{t^{2}} + 3 t^{2}$ ..... (i)

But $V = 0$ ...[Given]

$\begin{aligned} \Rightarrow \frac{- 48}{t^{2}} + 3 t^{2} = 0 \\ \Rightarrow t = 2 \end{aligned}$

Now,

$\begin{aligned} A & = \frac{d V}{d t} \\ = \frac{d}{d t} (\frac{- 48}{t^{2}} + 3 t^{2}) \\ = \frac{96}{t^{3}} + 6 t \end{aligned}$

Acceleration at $t = 2$ is $\frac{96}{8} + 12 = 24$

2. ⇒ (MHT CET 2021 20th September Morning Shift )

A particle is moving on a straight line. The distance $S$ travelled in time $t$ is given by $S = a t^{2} + b t + 6$ . If the particle comes to rest after 4 seconds at a distance of $16 m$ . from the starting point, then the acceleration of the particle is.

A. $\frac{- 3}{4} m / \sec^{2}$

B. $\frac{- 1}{2} m / \sec^{2}$

C. $- 1 m / \sec^{2}$

D. $\frac{- 5}{4} m / \sec^{2}$

Correct Option is (D)

S = at $^{2}$ + bt + 6

$V = \frac{dS}{dt} = 2 at + b$ and $A = \frac{d^{2} S}{{dt}^{2}} = 2 a$

When particle comes to rest,

S = 16, t = 4, V = 0

$∴$ 16 = a(4) $^{2}$ + b(4) + 6 $\Rightarrow$ 16a + 4b = 10 ..... (1)

Also 0 = 2a(4) + b $\Rightarrow$ b = $-$ 8a ..... (2)

From (1) and (2), we get

16a + 4(8a) = 10 $\Rightarrow$ $-$ 16a = 10 $\Rightarrow$ a = $\frac{- 5}{8}$

We have acceleration $A = 2 a = 2 (\frac{- 5}{8}) = \frac{- 5}{4}$ m/sec $^{2}$

⇒Application Of Derivative