8. the graph shows a velocity - time graph for a student moving in a straight line. from the graph: a)…

Explanation:

Step1: Recall acceleration formula

Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v = v_{f}-v_{i}$ and $\Delta t=t_{f}-t_{i}$.

Step2: For $t = 0\ s$ and $t = 3.0\ s$

From the graph, $v_{i}=0\ m/s$, $v_{f}=5\ m/s$, $\Delta t = 3.0 - 0=3.0\ s$. Then $a=\frac{5 - 0}{3.0}=\frac{5}{3}\approx1.67\ m/s^{2}$.

Step3: For $t = 3.0\ s$ and $t = 6.0\ s$

$v_{i}=5\ m/s$, $v_{f}=5\ m/s$, $\Delta t=6.0 - 3.0 = 3.0\ s$. Then $a=\frac{5 - 5}{3.0}=0\ m/s^{2}$.

Step4: For $t = 6.0\ s$ and $t = 9.0\ s$

$v_{i}=5\ m/s$, $v_{f}=- 5\ m/s$, $\Delta t=9.0 - 6.0 = 3.0\ s$. Then $a=\frac{-5 - 5}{3.0}=-\frac{10}{3}\approx - 3.33\ m/s^{2}$.

Step5: For $t = 9.0\ s$ and $t = 12.0\ s$

$v_{i}=-5\ m/s$, $v_{f}=-5\ m/s$, $\Delta t=12.0 - 9.0 = 3.0\ s$. Then $a=\frac{-5-(-5)}{3.0}=0\ m/s^{2}$.

Answer:

(i) $1.67\ m/s^{2}$ (ii) $0\ m/s^{2}$ (iii) $-3.33\ m/s^{2}$ (iv) $0\ m/s^{2}$