Question

Two motorcycles are travelling at speeds of 120 kilometers per hour and 90 kilometers per hour respectively. If the faster one travels for 4 hours longer than the slower one and also goes twice as far, how far did the slow one travel?

Answer 1

answer is 720km

Explanation:

hope this helped :)

Answer 2

The slower motorcycle traveled 720 km.

Explanation:

In order to obtain how far the slow one traveled, you have to use the equation
`v=(d)/(t)` (I)

where v is the speed,d is the distance and t is the time. Then, you have to use the given relations and solve system of equations.

Let v1 and v2 represent the speed of the faster one and the slower one, respectively.

Let d1,t1 and d2,t2 represent the distance and time traveled by the faster one and the slower one, respectively.

It is known that:

v1=120 km/h

v2=90km/h

t1=4+t2 (because the faster one travels for 4 hours longer than the slower one)

d1=2d2 (because the faster one goes twice as far as the slower one)

Using (I):

`t1=(d1)/(v1)` (II)

`t2=(d2)/(v2)` (III)

Replacing (II) and (III) in t1=4+t2

`(d1)/(v1)=4+(d2)/(v2)`

`(1)/(120)d1=4+(1)/(90)d2` (IV)

Replacing d1=2d2 in (IV):

`(1)/(120)(2)d2= 4+(1)/(90)d2\\(1)/(60)d2-(1)/(90)d2=4`

Using d2 as common factor:

`d2((1)/(60)-(1)/(90))=4`

Solving the fractions:

`d2((1)/(180))=4`

Multiplying by 180 both sides:

d2=180(4)=720 km