Question

On a drive from one city to​ another, Victor averaged 5151 mph. If he had been able to average 7575 ​mph, he would have reached his destination 88 hrs earlier. What is the driving distance between one city and the​ other?

Answer 1

Answer:

d=1.416.525 mile

Step-by-step explanation:

V1=5151m/h, t1=t, V2=7575m/h, t2=t-88h

d1=d2 Because it is same distance; V1=d1/t and V2=d2/(t-88) but d1=d2

d=V1t=V2(t-88) → 5151t=7575(t-88) → 5151t=7575t-666.600 → 7575t-5151t=666.600 → 2424t=666.600 → t=666.600/2424 → t=275h so

$d=5151\frac{mile}{h}.275h = 1.416.525mile$