Question

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 25.0 min at 95.0 km/h, 16.0 min at 100.0 km/h, and 50.0 min at 40.0 km/h and spends 30.0 min eating lunch and buying gas. Determine the average speed for the trip.

Answer 1

49.8 km/h

Step-by-step explanation:

We can determine the average speed by calculating the total distance covered and the time taken,

- First part of the trip:

v1 = 95.0 km/h = 26.4 m/s

t1 = 25.0 min = 1500 s

Distance covered:

`d_1 = v_1 t_1 = (26.4 m/s)(1500 s)= 39600 m = 39.6 km`

- Second part of the trip:

v2 = 100.0 km/h = 27.8 m/s

t2 = 16.0 min = 960 s

Distance covered:

`d_2 = v_2 t_2 = (27.8 m/s)(960 s)=26688 m= 26.7 km`

- Third part of the trip:

v3 = 40.0 km/h = 11.1 m/s

t3 = 50.0 min = 3000 s

Distance covered:

`d_3 = v_3 t_3 = (11.1 m/s)(3000 s)=33300 m = 33.3 km`

The person also spend 30.0 min without moving:

t4 = 30.0 min = 1800 s

Total distance covered:

`d=d_1 +d_2 +d_3 = 39.6 km +26.7 km + 33.3 km =99.6 km`

Total time taken:

`t=t_1 +t_2 +t_3+t_4 = 1500 s+ 960 s+ 3000 s+1800 s=7260 s = 2.0 h`

So, the average speed is

`v=(d)/(t)=(99.6 km)/(2.0 h)=49.8 km/h`