# Question 1f118

Jun 21, 2015

The answer is actually 5 hours.

#### Explanation:

The trick is to keep in mind that the question asks for the time required for car B to overtake car A.

The fact that car B starts 1 hour later than car A implies that car B must travel the same distance car A travelled, but it must do so one hour faster.

Let's say that car A travels a distance equal to $d$ in $t$ hours. This means that you can write

$d = {v}_{A} \cdot t$

Car B, on the other hand, must travel the same distance in $t - 1$ hours, so you have

$d = {v}_{B} \cdot \left(t - 1\right)$

This will get you

${v}_{A} \cdot t = {v}_{B} \cdot \left(t - 1\right)$

50cancel("km")/"h" * t = 60cancel("km")/"h" * (t-1)#

$50 t = 60 t - 60 \implies 10 t = 60 \implies t = \frac{60}{10} = \text{6 hours}$

However, this is the time needed for car A to reach distance d, which means that car B will need

${t}_{B} = {t}_{A} - 1 = 6 - 1 = \textcolor{g r e e n}{\text{5 hours}}$

to catch car A.

Jun 21, 2015

$t = 5 h$.

#### Explanation:

In this type of problem it is enjoying to use a trick.
In the instant in which the car B starts its motion, the car A is:

$s = v \cdot t = 50 \frac{k m}{h} \cdot 1 h = 50 k m$, far from B.

The trick consists in virtually stopping A, and giving to B the velocity:

$v = {v}_{B} - {v}_{A} = \left(60 - 50\right) \frac{k m}{h} = 10 \frac{k m}{h}$.

Now the question is different: we have to know in how many hours B can run $50 k m$ at the velocity $v = 10 \frac{k m}{h}$.

$s = v \cdot t \Rightarrow t = \frac{s}{v} = \frac{50 k m}{10 \frac{k m}{h}} = 5 h$.