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vodomira [7]
3 years ago
10

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"?
Physics
1 answer:
Ganezh [65]3 years ago
5 0

Answer:

230.4 s

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 = \frac{d}{v_A}=\frac{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 = \frac{d}{v_B}=\frac{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.

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