Trip to the Store
Your apartment and the grocery store are both located on the same side of Main Street and are separated by a distance $d=1\ km$. You decide to head out to pick up a few items despite the light rain.
1. To avoid getting too wet, you jog at constant speed to the store and get there in $5\ min$. What is your average velocity? What is your average speed?
View answerHide answerUsing $\hat{x}$ to indicate direction, your displacement vector from your apartment to the store is $\mathrm{\Delta }\overrightarrow{r}=1000\ \hat{x}$ (in meters). Thus, your average velocity is equal to
\boxed{{\overrightarrow{v}}_{avg}=\frac{\mathrm{\Delta }\overrightarrow{r}}{\mathrm{\Delta }t}=\frac{1000}{5\cdot 60}\ \hat{x}\approx 3.33\ \hat{x}\ \ \ \left(m/s\right)}
Your average speed is equal to the ratio of the total distance traveled over the total time and is therefore equal to
\boxed{v_{avg}=\frac{d}{\mathrm{\Delta }t}=\frac{1000}{5\cdot 60}\approx 3.33\ m/s}
2. After purchasing a couple of bags of groceries, you walk home. You initially walk quickly and leave the store with a speed $v_i=2\ m/s$. However, you gradually slow down as you walk toward your apartment and end up reaching it with a speed $v_f=1.2\ m/s$ after $20\ min$. What is your average acceleration?
View answerHide answerDuring your trip home, your velocity changes by an amount
\mathrm{\Delta }\overrightarrow{v}={\overrightarrow{v}}_f-{\overrightarrow{v}}_i=-2\ \hat{x}-\left(-1.2\ \hat{x}\right)=-0.8\ \hat{x}\ \ \ \ \ \left(m/s\right)
Your average acceleration is therefore equal to
\boxed{{\overrightarrow{a}}_{avg}=\frac{\mathrm{\Delta }\overrightarrow{v}}{\mathrm{\Delta }t}=-\frac{0.8}{20\cdot 60}\ \hat{x}\approx -6.67\times {10}^{-4}\ \hat{x}\ \ \left(m/s^2\right)}
3. What is your average velocity for the return trip? What is your average speed?
View answerHide answerYour displacement vector during your trip back home is equal to $\mathrm{\Delta }\overrightarrow{r}=-1000\ \hat{x}$ and therefore your average velocity is equal to
\boxed{{\overrightarrow{v}}_{avg}=\frac{\mathrm{\Delta }\overrightarrow{r}}{\mathrm{\Delta }t}=-\frac{1000}{20\cdot 60}\ \hat{x}\approx -0.83\ \hat{x}\ \ \ \left(m/s\right)}
Your average speed is equal to the ratio of the total distance traveled over the total time and is therefore equal to
\boxed{v_{avg}=\frac{d}{\mathrm{\Delta }t}=\frac{1000}{20\cdot 60}\approx 0.83\ m/s}
4. Overall, if you consider the entire trip from your apartment to the store and back, and assume that you spent $10\ min$ inside the store, what is your average velocity? What about your average speed? You may neglect any distance walked in the store to pick up your items in front of the distance between your apartment and the store.
View answerHide answerIf you consider the entire trip, your displacement vector is equal to $\mathrm{\Delta }\overrightarrow{r}=\overrightarrow{0}$ and therefore your average velocity is equal to
\boxed{{\overrightarrow{v}}_{avg}=\overrightarrow{0}\ \ \ \left(m/s\right)}
Your average speed, however, is not zero because it is the ratio of the total distance traveled over the total time spent. It is therefore equal to
\boxed{v_{avg}=\frac{2d}{\mathrm{\Delta }t_{tot}}=\frac{2\cdot 1000}{5\cdot 60+10\cdot 60+20\cdot 60}\approx 0.95\ \ m/s}