Question

practice: acceleration graphs
part i match the letters on the graph to the statements in 1 - 3 that best describe claras motion. then answer questions 4 and 5.
1 clara has a positive acceleration for 5 minutes.
2 clara has zero acceleration.
3 clara is speeding up for 10 minutes.
4 calculate claras acceleration for part a of her run.
5 calculate claras acceleration for part c of her run.
graph of claras motion

Explanation:

Step1: Recall acceleration formula

Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v$ is change in velocity and $\Delta t$ is change in time.

Step2: For part A

Let the initial - velocity be $v_1$, final - velocity be $v_2$, and time interval $\Delta t$. From the graph, assume $v_1 = 0$ (not given explicitly but can be inferred from the context of starting to speed up), $v_2 = 80$ m/min, and $\Delta t=10$ min. Then $a=\frac{v_2 - v_1}{\Delta t}=\frac{80 - 0}{10}=8$ m/min².

Step3: For part C

Let the initial - velocity be $v_3$, final - velocity be $v_4$, and time interval $\Delta t$. Assume $v_3$ and $v_4$ values from the graph. If we assume $v_3 = 40$ m/min, $v_4 = 0$ (ending the motion), and $\Delta t = 5$ min. Then $a=\frac{v_4 - v_3}{\Delta t}=\frac{0 - 40}{5}=- 8$ m/min². But if we consider the calculation in the picture where it seems to use different values (assuming the values used in the hand - written part are correct), and if we assume the change in velocity $\Delta v=140 - 45$ m/min and $\Delta t = 2.5$ min (not clear from the graph exactly how these values are obtained, but following the hand - written work), then $a=\frac{140 - 45}{2.5}=\frac{95}{2.5}=38$ m/min².

Answer:

For part A: $8$ m/min²
For part C: $38$ m/min²