(a) Powerful sports car can go from zero to 25m/s (about 60mi/h) in 5 s. (1) What is the magnitude of average acceleration? (2) How does this compare with the acceleration of a rock falling near the Earth’s surface? (b) Suppose the position of an object at time tis $\left(3+5t,4t^2,2t-6t^3\right)$ . (1) what is the instantaneous velocity at time t? (2) What is the instantaneous acceleration at time t? (3) What is the instantaneous velocity at time t=0? (4) What is the instantaneous acceleration at time t=0?

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, $∆\stackrel{\rightharpoonup }{v}=25m{s}^{-1}and∆t=5s.$

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

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

Therefore, the average acceleration is 5ms^-2.

(a)(2)

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

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

Therefore, the instantaneous velocity at time t is .

(b)(2)

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

Therefore, the instantaneous acceleration at time is .

(b)(3)

Substitute into the obtained instantaneous velocity.

Therefore, the instantaneous velocity at t=0 is .

(b)(4)

Substitute into the obtained instantaneous acceleration.

Thus, the instantaneous acceleration at t=0 is .