Question

A particle is moving on a straight line, where its position is a function of time $t$ given by $s=a{t}^2+bt+6\ge 0$. If it is known that the particle comes to rest after $4seconds$ at a distance of $16metres$ from the starting position $(t=0)$, then the retardation in its motion is

• A. $-1m/s^2$
• B. $\frac{5}{4}m/s^2$
• C. $\frac{1}{2}m/s^2$
• D. $-\frac{5}{4}m/s^2$

Answer 1

The correct option is B

$\frac{5}{4}m/s^2$

Step 1: Given data

Distance, $s=a{t}^2+bt+6$

Step 2: Find velocity, $v$

Velocity is the rate of change of displacement with respect to the time.

$s=a{t}^2+bt+6$

Differentiating with respect to the time

$$\begin{gathered} v=\frac{ds}{dt} \\ v=2at+beq\left(1\right) \end{gathered}$$

Step 3: Find acceleration, $A$

After $4seconds,v=0$ and distance $s=16m$

Putting these value in the above equation, we get the value of a

$$\begin{gathered} 0=2a\times 4+b \\ 8a+b=0 \\ s=a{t}^2+bt+6 \\ 16=16a+4b+6 \\ 16=16a+4(-8a)+6 \\ a=–\frac{5}{8} \end{gathered}$$

Acceleration is the rate of change of the velocity with respect to the time.

$$\begin{gathered} A=\frac{dv}{dt} \\ A=\frac{d}{dt}\left(2at+b\right) \\ A=2a \\ A=2\times \left(-\frac{5}{8}\right) \\ A=-\frac{5}{4}m{s}^{-2} \end{gathered}$$

Retardation $=\frac{5}{4}m{s}^{-2}$

Hence, option $\left(B\right)$ is correct.