Two cars travel westward along a straight highway, one at a constant velocity of 97 km/h, and the other at a constant velocity of 113 km/h. Assuming that both cars start at the same point, how much sooner does the faster car arrive at a destination 17 km away? Answer in units of h.

Respuesta :

Answer:

0.025hr

Explanation:

The full solution is on the image below. The two cars cover the same distance at different time intervals. Since the distance is constant, the velocity is inversely proportional to the time taken to cover the constant distance

Ver imagen pstnonsonjoku

Answer:

0.025 h

Explanation:

Let's assume for the first car, the destination is [tex]x_{1}[/tex], the time is [tex]t_{1}[/tex], the velocity is [tex]v_{1}[/tex] and for the second car the destination is [tex]x_{2}[/tex], the time is [tex]t_{2}[/tex], the velocity is [tex]v_{2}[/tex].

We are given:

[tex]v_{1}[/tex] = 97 km/h

[tex]v_{2}[/tex] = 113 km/h

If we are asked the time, the destinations must be equal which are also given:

[tex]x_{1}[/tex] = [tex]x_{2}[/tex] = 17

For constant velocity, the equation is x = v * t

Hence,

[tex]x_{1}[/tex] = [tex]v_{1}[/tex] * [tex]t_{1}[/tex] = [tex]x_{1}[/tex] = 97 * [tex]t_{1}[/tex] = 17

⇒ [tex]t_{1}[/tex] = 17/97 = 0.175 h

[tex]x_{2}[/tex] = [tex]v_{2}[/tex] * [tex]t_{2}[/tex] = [tex]x_{2}[/tex] = 113 * [tex]t_{2}[/tex] = 17

⇒ [tex]t_{2}[/tex] = 17/113 = 0.150 h

So,

[tex]t_{1}[/tex] -  [tex]t_{2}[/tex] = 0.175 - 0.150 = 0.025 h

The second car arrives 0.025 h sooner.

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