Question

Two cars, which start 200200200 miles apart, drive toward each other. The two cars move at a constant speed, but one car drives 151515 miles per hour faster than the other. After 222 hours, the cars pass each other. If rrr represents the speed of the slower car, which equation best models the situation?

Answer 1

To model the situation where two cars start 200 miles apart and drive towards each other at different speeds, we can use the equation d = rt, where d is the distance, r is the rate or speed, and t is the time.

Step-by-step explanation:

To model the situation where two cars start 200 miles apart and drive towards each other at different speeds, we can use the equation d = rt, where d is the distance, r is the rate or speed, and t is the time. Let's represent the speed of the slower car as r. The faster car's speed would be r + 15 miles per hour. After 2 hours, the distance covered by the slower car would be 2r, and the distance covered by the faster car would be 2(r + 15). The equation that models this scenario is:

2r + 2(r + 15) = 200

By simplifying this equation, we get:

2r + 2r + 30 = 200

4r + 30 = 200

4r = 170

r = 42.5

Answer 2

The equation that describes the situation is 200 = 4r + 30

Step-by-step explanation:

distance between the two cars = 200 miles

Let the speed of the slower car be r miles/hr

The speed if the faster car = (r + 15) miles/hr

The cars pass each other after 2 hours

Distance = rate * time

200 = (r*2) + (r +15)2

200 = 2r + 2r +30

200 = 4r + 30 (this is the equation that represents the situation)

Collect like terms

200 – 30 = 4r

170 = 4r

Therefore,

r = 170/4

r = 42.5 miles/hr for the slower car