ANSWER:
(i)
(a) 0 m/s²
(b) 1.6 m/s²
(c) 0.8 m/s²
(ii)
(a) 0 m/s²
(b) 1.6 m/s²
(c) 0 m/s²
STEP-BY-STEP EXPLANATION:
(i)
We have that the acceleration is given as follows:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]
(a) 0 s to 5 s:
[tex]a=\frac{v_5-v_0}{5-0}=\frac{-8-(-8)}{5-0}=\frac{-8+8}{5}=\frac{0}{5}=0\text{ }\frac{m}{s^2}[/tex]
(b) 5 s to 15 s:
[tex]a=\frac{v_{15}-v_5}{15-5}=\frac{8-(-8)}{15-5}=\frac{8+8}{10}=\frac{16}{10}=1.6\text{ }\frac{m}{s^2}[/tex]
(c) 0 s to 20 s:
[tex]a=\frac{v_{20}-v_0}{20-0}=\frac{8-(-8)}{20-0}=\frac{8+8}{20}=\frac{16}{20}=0.8\text{ }\frac{m}{s^2}[/tex]
(ii)
The instantaneous acceleration is given as follows:
[tex]a=\frac{dv}{\text{ dt}}\text{ (slope of graph)}[/tex]
(a) 2 sec (No slope so acceleration at 2 sec is 0 m/s²). We confirm it as follows:
[tex]a=\frac{v_2-v_0}{2-0}=\frac{-8-(-8)}{2-0}=\frac{-8+8}{2}=\frac{0}{2}=0\text{ }\frac{m}{s^2}[/tex]
(b) 10 sec
The slope is:
[tex]a=\frac{8-(-8)}{15-5}=\frac{8+8}{10}=\frac{16}{10}=1.6\text{ }\frac{m}{s^2}[/tex]
(c) 18 sec: No slope so acceleration at 18 sec is 0 m/s²
