Question

If the distance traveled by a point in time $\mathrm{t}$ is $\mathrm{s}=180\mathrm{t}–16{\mathrm{t}}^2$, then the rate of change in velocity is

• A. $–16\mathrm{t}$ unit
• B. $48$ unit
• C. $–32$ unit
• D. None of these

Answer 1

The correct option is C

$–32$ unit

Step 1 Given data:

The distance ‘$\mathrm{s}$’ traveled by a point in time $\mathrm{t}$ is $\mathrm{s}=180\mathrm{t}–16{\mathrm{t}}^2$

Step 2 Formula used:

The first derivative of displacement is velocity.

That is, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}$

Acceleration($\mathrm{a}$) is the derivative of velocity with respect to time.

That is, $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}$

Step 3 Velocity($\mathrm{v}$):

Velocity($\mathrm{v}$) is the rate of change in displacement of an object with respect to time.

Here, $\mathrm{s}=180\mathrm{t}–16{\mathrm{t}}^2$

Velocity, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}$

That is,

$$\begin{gathered} \mathrm{v}=\frac{d(180\mathrm{t}–16{\mathrm{t}}^2)}{\mathrm{dt}} \\ v=180-2\times 16t \\ v=180-32t \end{gathered}$$

Step 4 Acceleration($\mathrm{a}$):

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

Acceleration, $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}$

That is,

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{d}(180-32\mathrm{t})}{\mathrm{dt}} \\ \mathrm{a}=-32\mathrm{m}{\mathrm{s}}^{-2} \end{gathered}$$

The distance traveled by a point in time $\mathrm{t}$ is $\mathrm{s}=180\mathrm{t}–16{\mathrm{t}}^2$, then the rate of change in velocity is $–32\mathrm{m}{s}^{-2}$. Hence, option C is correct.