Question

Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 70 mi/h.

(a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 14 mi away?

(b) How far must the faster car travel before it has a 15-min lead on the slower car?

Answer 1

(a) The speed of faster car is 70 mil/h and speed of slower car is 55 mil /h

As we know,

Distance = Speed × Time

Time = Distance / Speed

To cover 14 miles,

Time for faster car(T1) = 14/70 = 0.2 h = 12 min

Time for slow car(T2) = 14/55 = 0.254 h 15 min

T2-T1 = 15 -12 3 min

So the faster car arrives approx 3 min earlier than slower one.

(b) Now for taking a 15 min lead means 0.25 h, let's assume it need to travel x distance.

So x/55 - x/70 = 0.25

15x/3850 = 0.25

x = (0.25 × 3850)/15

x = 64.16

x 64 miles

Hence the faster car need to travel approx 64 miles for getting a 15 min lead on slower one.