The question I'm solving is:
A student jogs from home to the store. She runs at 1 m/s for 1/3 of the time and 1.5 m/s for the remaining time. On her way back home she runs at 1 m/s for half the distance and at 1.5 m/s for the rest of the trip. Find her average speed for the round trip.
The way I approached this first was by defining:
- $d$ as the distance between the house and the store
- $t_1$ as the time to get there on the first leg
- $t_2$ as the time to get back
This means that the answer I would be looking for could be defined as $\frac{2d}{t_1+t_2}$
Setting
- $d=1\frac{m}{s}\cdot\frac{1}{3}t_1+1.5\frac{m}{s}\cdot\frac{2}{3}t_1$
- and $t_2=\frac{\frac{1}{2}d}{1\frac{m}{s}}+\frac{\frac{1}{2}d}{1.5\frac{m}{s}}$
I was able to simplify both down to
- $d=\frac{4}{3}\frac{m}{s}t_1$
- $t_2=\frac{ds}{2m}+\frac{ds}{3m}$
Solving for $t_2$ in terms of $t_1$ by substituting in my value for $d$, I yielded $t_2 = \frac{14}{3}t_1$.
Then, plugging all my values into $\frac{2d}{t_1+t_2}$, all units cancelled providing me with $v_{avg}=\frac{8}{17}m/s$.
This is apparently incorrect. However, I cannot find anything wrong with this approach.
The other option to tackling this problem is to calculate the average velocities of each individual leg of the trip.
By defining $v_1=\frac{1\frac{m}{s}\cdot\frac{1}{3}t_1+1.5\frac{m}{s}\cdot\frac{2}{3}t_1}{t_1}$ where $v_1$ is the velocity of the first leg of the trip, $v_1$ can be computed at $\frac{4}{3}\frac{m}{s}$.
The trip back can be divided into sections A and B, with variables $d$, $v$, and $t$ representing the distance, velocity, and time. Since $d_A=d_B$, $d_A=v_At_A$, and $d_B=v_Bt_B$,
$v_At_A=v_Bt_B$
$\therefore t_A=1.5t_B$
Solving for the velocity from here yields:
- $v_2=\frac{\Delta x}{\Delta t}$
- $v_2=\frac{v_At_A + v_Bt_B}{t_A+t_B}$
- $v_2=\frac{1\frac{m}{s}1.5t_B + 1.5\frac{m}{s}t_B}{1.5t_B+t_B}$
Simplifying gives $v_2 = \frac{6}{5}\frac{m}{s}$ or $1.20\frac{m}{s}$
Averaging $v_1$ and $v_2$ yields $1.26\frac{m}{s}$, a wildly different answer then the about $0.47\frac{m}{s}$ from the first method.
I am told that the answer yielded by the second method is correct, yet I cannot find any problems with the first. Can someone help with this?
Found it. Made an algebra mistake. In the first method, when substituting $d$ in for $t_2$ to solve for $t_2$ in terms of $t_1$, I should have gotten $t_2=\frac{10}{9}t_1$. That then provides the correct answer.