# The average velocity is closer to the low velocity than to the higher velocity

Bob bikes from his home against the wind with a constant velocity of 12 km/h. Then he bikes back to his home with the wind with a higher constant velocity. His average velocity during the trip is 16 km/h.

What was his velocity while biking back home?

My attempt:

I would say that his velocity while biking back home was 20 km/h since: $$v_{avg}=\frac{v_1+v_2}{2}\\ 16=\frac{12+v_2}{2}\\ 32=12+v_2\\ v_2=20$$

However the book says the solution is 24 km/h. It also says that the average velocity is closer to the low velocity than to the higher velocity. What do they mean by that?

• when the displacement is 0, how can the average velocity be 16 km/h ?
– Our
Oct 7, 2016 at 10:34
• @Leth ikr it must be avg speed Oct 7, 2016 at 10:36
• Then it must be written as average speed, or else my comment is the answer of this question.
– Our
Oct 7, 2016 at 10:38

First of all, it would be average speed not velocity b/c average velocity would be zero due to zero displacement [ Be clear with concepts first, Avg velocity = Total displacement / total time

Avg speed = Total distance / total time

I hope you know the difference b/w dist. And disp. ]

Now, Let speed for way back be v let distance travelled in one side = d Total dist. travelled = 2d Time taken for one side = d/12 Time taken for second trip = d/v Total time = d/v + d/12 Avg speed = total distance / total time = 2d ÷ ( d/v + d/12) 16 = (2 ÷ 1/v + 1/12) Solving , v = 24 It can easily be seen avg speed ( 16} is closer to lower speed that is 12 km/hr

When travelling slower it takes more time and so has a greater effect on the average speed than when travelling faster.
Think of travelling one way at 1 km/hr and the other way at 100 km/hr the distance travelled is the same both ways but the total time of travel is dominated by the slower speed and so in this case the expected average speed would be closer to 1 km/hr than 100 km/hr.

To solve your problem find the times to go one way and then the other and hence the total time of travel and then divide this into the total distance travelled.
This will give you the average speed.

From this you can find the speed on the return journey.

• Thanks for the answer. Would you like to add some calculations? I understand that in respect to a fixed distance a high speed would result in less time taken than lower speed. What I don't know is how to explain it in mathematical terms. Oct 7, 2016 at 10:27
• Call the one way distance $d$ and the speed of return $v$. Get an equation for the total time taken going there and coming back in terms of $d$ and $v$ . Divide this time into the total distance travelled and equate the result to the given average speed. Oct 7, 2016 at 10:34
• @Fracher Thanks, it was sufficient for me to realize that since the denominators aren't equal $\dfrac{v_1+v_2}{2}$ does not apply in this case. It's the first time I am solving this equation so I want to do everything step-by-step at least the first few times. Oct 7, 2016 at 12:12

$$\begin{array}{lll}\\ v_{avg-neq.v}&=\dfrac{v_1+v_2}{2} &/ v=\frac{\Delta x}{\Delta t}\\\\ &=\dfrac{\dfrac{\Delta x_1}{\Delta t_1}+\dfrac{\Delta x_2}{\Delta t_2}}{2} & /\Delta t_1\neq \Delta t_2; \Delta x_1=\Delta x_2\\\\ &=\dfrac{\dfrac{\Delta x_1\Delta t_2}{\Delta t_1\Delta t_2}+\dfrac{\Delta x_1\Delta t_1}{\Delta t_2\Delta t_1}}{2}\\\\ &=\dfrac{\dfrac{\Delta x_1(\Delta t_2+\Delta t_1)}{\Delta t_1\Delta t_2}}{2} & /\dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{ad}{bc} \text{ (mtp.2)}\\\\ &=\dfrac{\Delta x_1(\Delta t_2+\Delta t_1)}{2\Delta t_1\Delta t_2}\\\\\\ \Delta x_t=v_{avg}t_t & /t_t=t_1+t_2\\ \Delta x_t=v_{avg}(t_1+t_2) &/v_{avg}=16\\ \Delta x_t=16(t_1+t_2) & /\div 2\\ \frac{1}{2}\Delta x_t=8t_1+8t_2 &/\frac{1}{2}\Delta x_t=\Delta x_1\\ \Delta x_1=8t_1+8t_2 &/\Delta x_1=12t_1\\ 12t_1=8t_1+8t_2 &/-8t_1\\ 4t_1=8t_2 &/\div 4\\ t_1=2t_2 & /\Delta x_{1,2}=vt\\ 12t_1=\frac{1}{2}v_2t_1 &/\div t_1\\ 12=\frac{1}{2}v_2 &/\times 2\\ v_2=24 \end{array}$$