Question

1. A train travels 20 km at a uniform speed of 60 kmph and the next 20 km at a uniform speed of 80 kmph. Calculate the average speed?

2. A bus travels 40 km at a speed of 50 kmph and next 4 m at a speed of 30 kmph. Calculate its average speed?

Answer 1

Task.

1. A train travels $s_{1}=20$ km at a uniform speed of $v=60$ kmph and the next $s_{2}=20$ km at a uniform speed of $v_{2}=80$ kmph. Calculate the average speed.

2. A bus travels $s_{1}=40$ km at a speed of $v_{1}=50$ kmph and next $s_{2}=4$ m at a speed of $v_{2}=30$ kmph. Calculate its average speed.

Solution. 1) Let $t_{1}$ and $t_{2}$ be the times when train moved by the first and by the second parts of the path.

Thus

$t_{1}=\frac{s_{1}}{v_{1}},\qquad t_{2}=\frac{s_{2}}{v_{2}}.$

Then by definition the average speed is equal to

$v_{av}=\frac{s_{1}+s_{2}}{t_{1}+t_{2}}.$

Substituting expressions for $t_{1}$ and $t_{2}$ into the latter formula we will get the formula for average speed:

$v_{av}=\frac{v_{1}t_{1}+v_{2}t_{2}}{t_{1}+t_{2}}=\frac{s_{1}+s_{2}}{\frac{s_{1}}{v_{1}}+\frac{s_{2}}{v_{2}}}=\frac{(s_{1}+s_{2})v_{1}v_{2}}{s_{1}v_{2}+v_{1}s_{2}}.$

Now substitute the values $s_{1}$, $s_{2}$, $v_{1}$ and $v_{2}$:

$v_{av}=\frac{(20+20)*60*80}{20*80+60*20}=\frac{192000}{2800}=68.57\ kmph.$

2) Problem 2 is similar to 1) and differs only by values of distances and velocities. In particular, we will get the same fomula for the average velocity:

$v_{av}=\frac{(s_{1}+s_{2})v_{1}v_{2}}{s_{1}v_{2}+v_{1}s_{2}}.$

Substituting values we get:

$v_{av}=\frac{(40+4)*50*30}{40*30+50*40}=\frac{66000}{3200}=20.63\ kmph.$