Question

the displacement (in meters) of a particle moving in a straight line is given by the equation of motion $s = \frac{4}{t^{2}}$, where $t$ is measured in seconds. find the velocity (in m/s) of the particle at times $t = a$, $t = 1$, $t = 2$, and $t = 3$.

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v=\frac{ds}{dt}$, and $s = \frac{4}{t^{2}}=4t^{- 2}$.

Step2: Differentiate $s$ with respect to $t$

Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v=\frac{d}{dt}(4t^{-2})=4\times(-2)t^{-2 - 1}=-8t^{-3}=-\frac{8}{t^{3}}$.

Step3: Find velocity at $t = a$

Substitute $t = a$ into $v$, we get $v=-\frac{8}{a^{3}}$ m/s.

Step4: Find velocity at $t = 1$

Substitute $t = 1$ into $v$, we get $v=-\frac{8}{1^{3}}=-8$ m/s.

Step5: Find velocity at $t = 2$

Substitute $t = 2$ into $v$, we get $v=-\frac{8}{2^{3}}=-1$ m/s.

Step6: Find velocity at $t = 3$

Substitute $t = 3$ into $v$, we get $v=-\frac{8}{3^{3}}=-\frac{8}{27}$ m/s.

Answer:

$t = a$: $-\frac{8}{a^{3}}$ m/s
$t = 1$: $-8$ m/s
$t = 2$: $-1$ m/s
$t = 3$: $-\frac{8}{27}$ m/s