Suppose that during a test drive of two cars, one car travels 224 miles in the same time that a second car travels 160 miles. If the speed of the first car is 16 miles per hour faster than the speed of the second car, find the speed of both cars.

Speed of first car= 48 mph

Speed of second car= 32 mph

Explanation:

From the question, the following parameters are given

Distance travelled by the first car (d1) =224 miles

Distance travelled by the second car (d2)= 160 miles (it is obvious the second car is not as fast as the first car) since both cars attain these distances at the same time.

Time for the first car to attain 224 miles (t1)

Time for the second car to attain 160miles (t2). From the question, t1=t2(i)

Speed of the first car is 16mph faster than the second car. Then speed of first car

(s1 =s2+16)(ii)

Where s1 is speed of first car and s2 is speed of second car.

Speed=distance/time travelled

For the first car, s1=d1/t1(iii)

Substituting (ii) and d1=224 miles into (iii), we have

s2+16=224/t1

Making t1 the subject of the formular

t1=244/(s2+16)(iv)

For the second car, s2=d2/t2...(v)

Substituting d2=160 miles into (v),

We have

s2=160/t2

Making t2 the subject of the formular

t2=160/s2(vi)

Since t1=t2, we can equate (iv) and (vi) to become

224/(s2+16)=160/s2

By cross multiplying, you have

224s2=160s2+2560

You will obtain s2=32mph when you solve the equation above.

Since s1=s2+16

s1=32+16=48mph

im guessing the west of station b

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