Question

It’s asking to find the instantaneous acceleration at 10 seconds

Answer 1

The average acceleration is 0 m/s^2 during intervals of constant velocity, and the instantaneous acceleration at 10 seconds is 1.6 m/s^2, derived from the velocity function a(t) = 1.6 m/s^2.

i) Average Acceleration:

a) 0 seconds to 5 seconds: Since the velocity is constant, the average acceleration is 0 m/s^2.

b) 5.0 seconds to 15 seconds: Since the velocity is constant, the average acceleration is 0 m/s^2.

c) 0 seconds to 20 seconds: If the velocity is constant over this entire interval, the average acceleration is 0 m/s^2.

ii) Instantaneous Acceleration:

a) At 2.0 seconds: Since the velocity is constant, the instantaneous acceleration is 0 m/s^2.

b) At 10 seconds: To find the instantaneous acceleration at t = 10 seconds, you need to find the derivative of the velocity function with respect to time t:

Given that v(t) = 1.6t - 16,

The acceleration function a(t) is the derivative of v(t) with respect to t:

The formula is a(t) = dv/dt = 1.6 m/s^2.

So, the correct answer for the instantaneous acceleration at 10 seconds is 1.6 m/s^2, not 0 m/s^2.

Answer 2

Note:

At constant velocity, the acceleration is zero.

Acceleration is the change in velocity with time

i) Find the average acceleration of this object during the following time intervals

a) 0 s to 5 s

Since the velocity is constant at 0s to 5s, average acceleration = 0m/s²

b) 5.0 s to 15 s

`\begin{gathered} a=(v_2-v_1)/(t_2-t_1) \\ a=(8-(-8))/(15-5) \\ a=(16)/(10) \\ a=1.6\text{ m/s}^(2) \end{gathered}`

c) 0 s to 20 s

`\begin{gathered} a=(v_2-v_1)/(t_2-t_1) \\ a=(8-(-8))/(20-0) \\ a=(16)/(20) \\ a=0.8\text{ m/s}^(2) \end{gathered}`

ii) Find the instantaneous at the following times

a) 2.0 s

Since the velocity is contant at 2.0s, the instantaneous acceleration is 0 m/s²

b) 10 s

To find the equation represented by the line from 5 s to 15 s, select two points on the line

(5, -8) and (15, 8)

`\begin{gathered} \text{The slope, m = }(v_2-v_1)/(t_2-t_1) \\ m=(8-(-8))/(15-5) \\ m=(16)/(10) \\ m=1.6\text{ m/s}^(2) \end{gathered}`
`\begin{gathered} v-v_1=m(t-t_1) \\ v-(-8)=1.6(t-5) \\ v+8=1.6t-8 \\ v=1.6t-8-8 \\ v=1.6t-16 \end{gathered}`

The instantaneous velocity is:

v(t) = 1.6t - 16

`\begin{gathered} a(t)=(dv)/(dt) \\ a(t)=1.6 \\ a(10)=1.6\text{ m/s}^(2) \end{gathered}`

Instantaneous acceleration at 10 seconds = 1.6 m/s²