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 60 mi/h.

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

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

Answer 1

Solution:

Given:

Speed of first car (v₁) = 55 mi/h

Speed of second car (v₂) = 60 mi/h

and, Distance (d) = 14 mi

Now: Speed = distance / time

Thus: time taken = distance / speed.

Therefore: time taken by first car to travel 14 miles = t₁ = (14 mi) / (55 mi/h) = 0.2545 hours = 15.27 minutes

and, time taken by second car to travel 14 miles = t₂ = (14 mi) / (60 mi/h) = 0.233 hours = 14 minutes

Hence, Second car will reach (15.27 min - 14 min) = 1.27 minutes earlier than the first car.