Question

figure shows the motion - diagram for a horse galloping in one direction along a straight path. not every dot is labeled, but the dots are at equally spaced instants of time. part a what is the horses velocity during the first 10 seconds of its gallop? express your answer with the appropriate units. part b what is the horses velocity during the interval from 30 s to 40 s? express your answer with the appropriate units.

Explanation:

Step1: Recall velocity formula

The formula for average velocity is $v=\frac{\Delta x}{\Delta t}$, where $\Delta x$ is the change in position and $\Delta t$ is the change in time.

Step2: Solve for Part A

For the first 10 - s of its gallop, at $t = 0\ s$, assume $x_0=0\ m$ (not shown but can be assumed for the start - point), at $t = 10\ s$, $x_1 = 650\ m$. Then $\Delta x=x_1 - x_0=650 - 0=650\ m$ and $\Delta t = 10\ s$. So $v_A=\frac{\Delta x}{\Delta t}=\frac{650\ m}{10\ s}=65\ m/s$.

Step3: Solve for Part B

We need to estimate the positions at $t = 30\ s$ and $t = 40\ s$. At $t = 30\ s$, $x_{30}=350\ m$. To estimate the position at $t = 40\ s$, we note the motion is somewhat linear between $30\ s$ and $50\ s$. The change in position from $30\ s$ to $50\ s$ is $\Delta x'=250 - 350=- 100\ m$ in $\Delta t'=20\ s$. The average velocity in the $30 - 50\ s$ interval is $v'=\frac{-100\ m}{20\ s}=- 5\ m/s$. Assuming constant - velocity motion in the $30 - 40\ s$ interval (a reasonable estimate), $\Delta t_{30 - 40}=10\ s$, and $\Delta x_{30 - 40}=v'\times\Delta t_{30 - 40}=-5\ m/s\times10\ s=-50\ m$. So the position at $t = 40\ s$ is $x_{40}=350-50 = 300\ m$. Then $\Delta x=x_{40}-x_{30}=300 - 350=-50\ m$ and $\Delta t = 10\ s$. So $v_B=\frac{\Delta x}{\Delta t}=\frac{-50\ m}{10\ s}=-5\ m/s$.

Answer:

Part A:
Value: 65
Units: m/s
Part B:
Value: - 5
Units: m/s