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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"?

Respuesta :

skyluke89

Answer:

230.4 s

Explanation:

The speed of car A is

[tex]v_A = 35 mi/h[/tex]

and the distance travelled is

[tex]d = 10 mi[/tex]

so the time taken for car A is

[tex]t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h[/tex]

The speed of car B is

[tex]v_B = 45 mi/h[/tex]

and the distance travelled is

[tex]d = 10 mi[/tex]

so the time taken for car B is

[tex]t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h[/tex]

So the difference in time is

[tex]\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h[/tex]

Which corresponds to

[tex]\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s[/tex]

so car B arrived 230.4 s before car A.

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