# Thread: Need Help Solving A Problem

1. ## Need Help Solving A Problem

The time it takes Eric to drive to Marta's place varies inversely with his driving rate. When he drives at 30 mph, it takes him 45 minutes. If he makes the drive in half an hour, how fast is he going?

2. Originally Posted by dperks91
The time it takes Eric to drive to Marta's place varies inversely with his driving rate. When he drives at 30 mph, it takes him 45 minutes. If he makes the drive in half an hour, how fast is he going?
Hi dperks91,

if there is an inverse relationship,

then their product is a constant.

$av. speed=\frac{distance}{time}$

As distance is constant here since he travels the same journey in both instances

$(av. speed)(time)=distance$

$30\frac{45}{60}=distance$

if one is 30, the other is 45,

hence if he takes 30 minutes, the speed must be 45 km/hr

$45\frac{30}{60}=distance$

Alternatively, an inverse relationship may be written

$s=\frac{k}{t}$

$st=k$

$k=30(45)=1350$

$s=\frac{1350}{30}=45$

Yet another way is...

He travels for 45 minutes at 30km/hr
45 minutes is 3/4 of an hour

The distance travelled is $\frac{3}{4}30=\frac{90}{4}$ km

In 30 minutes, which is half an hour, he travels that distance.

He would have travelled twice that distance in 1 hour.

That's a speed of $\frac{90}{2}$ km/hr, hence his speed is 45 km/hr

3. Originally Posted by dperks91
The time it takes Eric to drive to Marta's place varies inversely with his driving rate. When he drives at 30 mph, it takes him 45 minutes. If he makes the drive in half an hour, how fast is he going?
Hi dperks91

Here's another approach.

Time varies inversely as rate: $t=\frac{k}{r}$

$.75=\frac{k}{30}$

$k=22.5$

This is your constant of variation. Apply this to the new situation.

$.5=\frac{22.5}{r}$

$r=45$

Just be sure your time is written in hours since speed is in miles per hour.