Question

A person takes a trip, driving with a constant speed of 92.5 km/h, except for a 28.0-min rest stop. The person's average speed is 72.2 km/h. How much time is spent on the trip?

Answer 1

40.3 min

Step-by-step explanation:

First of all, let's convert every quantity into SI units:

`v_1 = 92.5 km/h = 25.7 m/s` (speed in the first part of the trip)

`t_2 = 28.0 min = 1680 s` time during which the person has stopped

`v=72.2 km/h = 20.1 m/s` (average speed of the whole trip)

The average speed is the ratio between the total distance covered, d, and the total time taken, t:

`v=(d)/(t)` (1)

The total distance covered is simply

`d = v_1 t_1`

where
`t_1` is the time during which the person has moved at 92.5 km/h.

The total time taken is

`t= t_1 + t_2`

So (1) becomes

`v=(v_1 t_1)/(t_1 + t_2)`

Solving for
`t_1`:

`v t_1 + v t_2 = v_1 t_1\\vt_2 = (v_1+v)t_1\\t_1 = (v t_2)/(v_1+v)=((20.1 m/s)(1680 s))/(25.7 m/s + 20.1 m/s)=737.3 s`

which corresponds to

`t_2 = 737.3 s = 12.3 min`

So the total time of the trip is

`t = 28.0 min + 12.3 min = 40.3 min`