Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 65 mi/h. (a) assuming they start at the same point, how much sooner does the faster car arrive at a destination 15 mi away? (b) how far must the faster car travel before it has a 15-min lead on the slower car?

Let us say that:

A = slower car at 55mi/h

B = faster car at 65mi/h

1. Calculate time travelled of each car

A = 15 mi / (55mi/h) = 0.27 h

B = 15 mi / (65mi/h) = 0.23 h

Therefore:

0.23 – 0.27 = 0.04 h = 2.52 minutes

The faster car arrived sooner at 0.04 hours or 2.52 minutes

2. To get a 15 minute lead or 0.25h lead:

(x / 65) – (x / 55) = -0.25

x = -0.25 / (1/65 – 1/55)

x = 89.375 miles

The faster car has to travel 89.375 miles to have 15-min lead.

multiply 2kg by latent heat of vap

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