Question

Question #1f118

Answer 1

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 * t`

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

`d = v_B * (t-1)`

This will get you

`v_A * t = v_B * (t-1)`

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

`50t = 60t-60 => 10t = 60 => t = 60/10 = "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 = color(green)("5 hours")`

to catch car A.

Answer 2

`t=5h`.

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*t=50(km)/h*1h=50km`, far from B.

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

`v=v_B-v_A=(60-50)(km)/h=10(km)/h`.

Now the question is different: we have to know in how many hours B can run `50km` at the velocity `v=10(km)/h`.

The answer is easy:

`s=v*trArrt=s/v=(50km)/(10(km)/h)=5h`.