Question

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2.4 h, and Car B traveled the distance in 4 h. Car B traveled 22 mph faster than Car A. How fast did Car A travel? (The formula R⋅T=D , where R is the rate of speed, T is the time, and D is the distance can be used.) Enter your answer in the box.

Answer 1

Answer:

Car A traveled at 55 mph.

Step-by-step explanation:

Suppose, the speed of Car A is  $x$ mph.

As Car B traveled 22 mph faster than Car A , so the speed of Car B will be: $(x+22) mph$

Given that, Car A traveled the distance in 2.4 hours and Car B traveled the distance in 4 hours.

As,  $Distance= Speed*Time$

So, the distance traveled by Car A $=2.4x$ miles

and the distance traveled by Car B $=4(x+22)$ miles

Given that, two cars traveled equal distances. So, the equation will be......

$2.4x=4(x+22)\\ \\ 2.4x=4x+88\\ \\ 2.4x-4x=52.8\\ \\ -1.6x=52.8\\ \\ x=\frac{52.8}{-1.6}=-33$

Thus, the speed of Car A will be -33 mph.

Answer 2

Answer:

The speed of Car A = -55 km/h

Step-by-step explanation:

Given, Time taken by Car A = 2.4 hour

and Time taken by Car B = 4 hour

Also, Car B traveled 22 mph faster than Car A

Let the speed of Car A = x km/h

Then the speed of Car B = (x + 22) km/h

As we know that Distance = Speed × Time

⇒ Distance of Car A = 2.4 × x = 2.4x km

and Distance of Car B = 4 × (x + 22) = 4(x + 22) km

Also it is given that distance traveled by both the car is same.

⇒ 2.4x = 4(x + 22)

⇒ 2.4x = 4x + 88

⇒ 2.4x - 4x = 88

⇒ -1.6x = 88

⇒ x = -55

Thus, the speed of Car A = -55 km/h

and, the speed of Car B = -55 + 22 = -33km/h