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
[tex]\begin{gathered} a=\frac{v_2-v_1}{t_2-t_1} \\ a=\frac{8-(-8)}{15-5} \\ a=\frac{16}{10} \\ a=1.6\text{ m/s}^{2} \end{gathered}[/tex]
c) 0 s to 20 s
[tex]\begin{gathered} a=\frac{v_2-v_1}{t_2-t_1} \\ a=\frac{8-(-8)}{20-0} \\ a=\frac{16}{20} \\ a=0.8\text{ m/s}^{2} \end{gathered}[/tex]
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)
[tex]\begin{gathered} \text{The slope, m = }\frac{v_2-v_1}{t_2-t_1} \\ m=\frac{8-(-8)}{15-5} \\ m=\frac{16}{10} \\ m=1.6\text{ m/s}^{2} \end{gathered}[/tex][tex]\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}[/tex]
The instantaneous velocity is:
v(t) = 1.6t - 16
[tex]\begin{gathered} a(t)=\frac{dv}{dt} \\ a(t)=1.6 \\ a(10)=1.6\text{ m/s}^{2} \end{gathered}[/tex]
Instantaneous acceleration at 10 seconds = 1.6 m/s²
