91Ó°ÊÓ

The average acceleration during a certain interval is defined as the change in velocity for that interval. The average acceleration, unlike acceleration, is determined for a certain interval.

(a)(1)

From the given information, $\mathbf{∆}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}\mathbf{=}\mathbf{}\mathbf{25}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∆}\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{}\mathbf{s}$

Divide change in velocity by change in time to get the average acceleration.

$$\begin{gathered} {\stackrel{â‡\P Ä}{a}}_{avg}=\frac{∆\stackrel{â‡\P Ä}{v}}{∆t} \\ =\frac{25}{5} \\ =5m{s}^{-2} \end{gathered}$$

Therefore, the average acceleration is 5 $m{s}^{-2}$.

(a)(2)

The acceleration of a rock falling near the Earth’s surface is $9.8m{s}^{-2}$. While, in this case, the acceleration of the car is $5m{s}^{-2}$.

Therefore, the acceleration of a rock falling near the Earth’s surface is greater than the acceleration of the car.

(b)(1)

The given vector is t =.

Take the derivative of the given vector to compute the Instantaneous velocity.

Therefore, the instantaneous velocity at the time t is .

(b)(2)

Take the derivative of the instantaneous velocity to compute the Instantaneous acceleration.

Therefore, the instantaneous acceleration at the t is

(b)(3)

Substitute t=0 into the obtained instantaneous velocity.

Therefore, the instantaneous velocity at t = 0 is localid="1656669142087" .

(b)(4)

Substitute into the obtained instantaneous acceleration.

Thus, the instantaneous acceleration at is .