Question

In Speed Study Number 1, we looked at two cars traveling the same distance at different speeds on city streets. Car "A" traveled at a speed of 35 miles per hour. Car "B" traveled at 45 miles per hour. Both cars traveled a distance of 10 miles. How much sooner did Car "B" arrive at its destination than Car "A"?

Answer 1

230.4 s

Step-by-step explanation:

The speed of car A is

`v_A = 35 mi/h`

and the distance travelled is

`d = 10 mi`

so the time taken for car A is

`t_A = (d)/(v_A)=(10 mi)/(35 mi/h)=0.286 h`

The speed of car B is

`v_B = 45 mi/h`

and the distance travelled is

`d = 10 mi`

so the time taken for car B is

`t_B = (d)/(v_B)=(10 mi)/(45 mi/h)=0.222 h`

So the difference in time is

`\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h`

Which corresponds to

`\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s`

so car B arrived 230.4 s before car A.