Question

While driving home from school you travel at 95 km/h for 130 km then slow to 65 km/h. You get home in 3 hours and 20 min. How far is your hometown from school and what is the average speed?

Answer 1

Task. While driving home from school you travel at $v_{1}=95$ km/h for $d_{1}=130$ km then slow to $v_{2}=65$ km/h. You get home in $t=3$ hours and 20 min. How far is your hometown from school and what is the average speed?

Solution. Let $t_{1}$ be the time during which I travelled with velocity $v_{1}$ for the distance $d_{1}$, $t_{2}$ be the time when I travelled the rest of the distance with velocity $v_{2}$. Then

$t=t_{1}+t_{2},\qquad t_{1}=\frac{d_{1}}{v_{1}}.$

The distance which I travelled with velocity $v_{2}$ is equal to

$d_{2}=v_{2}t_{2}.$

We should find

$d=d_{1}+d_{2},$

and the average velocity $v$

$v=\frac{d_{1}+d_{2}}{t_{1}+t_{2}}=\frac{d}{t}.$

From the above formulas it follows that

$t_{2}=t-t_{1}=t-\frac{d_{1}}{v_{1}}$

$d_{2}=v_{2}t_{2}=v_{2}\left(t-\frac{d_{1}}{v_{1}}\right)=tv_{2}-d_{1}v_{2}/v_{1}.$

Hence

$d=d_{1}+d_{2}=d_{1}+tv_{2}-d_{1}v_{2}/v_{1}=tv_{2}+d_{1}(1-v_{2}/v_{1})=tv_{2}+\frac{d_{1}(v_{1}-v_{2})}{v_{1}}$

$v=\frac{d}{t}=\frac{tv_{2}+\frac{d_{1}(v_{1}-v_{2})}{v_{1}}}{t}=v_{2}+\frac{d_{1}(v_{1}-v_{2})}{tv_{1}}.$

Substituting values we get

$t=3$ hours and 20 min $=3\frac{1}{3}=\frac{10}{3}$ hours

$d=tv_{2}-\frac{d_{1}(v_{1}-v_{2})}{v_{1}}=\frac{10}{3}\cdot 65+\frac{130\cdot(95-65)}{95}=\frac{650}{3}+\frac{3900}{95}=$

$=216\frac{2}{3}+41\frac{1}{19}=257\frac{38+3}{3\cdot 19}=257\frac{41}{57}\approx 260\ km,$

$v=\frac{d}{t}=\frac{257\frac{41}{57}}{10/3}=\frac{14690/57}{10/3}=\frac{14690\cdot 3}{57\cdot 10}=\frac{1469}{19}=77\frac{6}{19}\approx 77\ km/hour.$

Answer. $d\approx 260\ km,\ v=77\ km/hour.$