Word Problem?

Question

One car travels 200 mi. A second car, traveling 10 mi/h faster than the first car, makes the same trip in 1 hour less time. Find the speed of the slower car.

Answer 1

Let car 1's speed = x. Then car 1 took 200/x hours to go 200 miles.

Car 2's speed = x+10
Car 2's time = 200/x - 1

For car 2, we have

Distance = time * speed
→ 200 = (200/x - 1) * (x+10).

Solve for x.

Answer 2

So we get: 200 = v*t
and 200 = (v+10)*(t-1)
(with v = speed of the slow car, t = time the slow car needs in hour).

=> 200 = v*t <=> 200/v=t
and 200 = v*t+10t -v -10

=> 200 = v*t
and 200 = v*(200/v) +10(200/v) -v -10

=> 200 = v*t
and 200 = 200 +2000/v -v -10 <=> 0 = 2000 -10v - v²
=> v = -5+ sqrt(25+2000) (the other solution is < 0, what makes no sense in this situation).
=> v = 40 mi/h !


(=> t = 200/v = 5h.
Test: 50 mi/h Car needs 4 ours for 200 mi #).

Answer 3

The slower car is doing 40 mph and does the trip in 5 hours. The faster car at 50 mph is doing the trip in 4 hours.