What Speed is Needed to Reach the Airport on Time After Traffic Delay?

[darklich21]

Homework Statement

You allow yourself 35 minutes to drive 25 miles to the airport, but are caught in heavy traffic and average 22 miles/hour for the first 15 minutes. What must your average speed be on the rest of the trip if you are to get there on time? Express your answer in miles/hour.

Homework Equations

D=vt

The Attempt at a Solution

This was confusing, but this is how i started. I did a little dimensional analysis and did (25mil/35min) x (60min/1hr)= 42.857 mil/hr. From what I understand, this is what you need to do to get to the airport on time. I'm going 22miles/hour for the first 15 mins, meaning, i only have (35-15) = 20 mins left to get to the airport on time. Then I did 42.857-the miles i orginally traveled, which in this case was 22, leaving me with 20.857 miles to go.

So I used D=vt.
20.857=(v)(20min). V=1.04mil/hour

This would be the speed I would need to be traveling for the next 20 mins in order to get to the airport on time.

Is this correct?

 

[rl.bhat]

In the first 15 minutes the average velocity is 22 miles/h. Let v be the average velocity in the remaining time.
So the net average velocity = (22+v)/2

 

[darklich21]

Hmm, If I'm solving for V, but how do I get that "net average velocity"?

 

[rl.bhat]

Hmm, If I'm solving for V, but how do I get that "net average velocity"?

You have already calculated it. That is total distance /total time.

 

[darklich21]

So if I am understanding this right, total distance/total time was 25mil/35 min?, which was 42.857mi/hr? and I am solving for V?

if so, I did this: 42.857= (22+v)/2
v=63.7mi/hr, which would be the speed I would need to go in order to make it to the airport on time.

Is this it?

 

[rl.bhat]

So if I am understanding this right, total distance/total time was 25mil/35 min?, which was 42.857mi/hr? and I am solving for V?

if so, I did this: 42.857= (22+v)/2
v=63.7mi/hr, which would be the speed I would need to go in order to make it to the airport on time.

Is this it?

I think so.

 

[darklich21]

ouch, apparently it was wrong lol.

I dunno, I think I'm supposed to use the "22mil/hour for 15 min" data somehow, perhaps a bit of conversion

any other ideas?

 

[rl.bhat]

Try this one.
In 15 minutes he moves 5.5 miles
Find the remaining distance and remaining time. Then find the average velocity for the second part.

 

[darklich21]

How did you get 5.5 kilometers? Did you use a conversion factor between kilomters and miles from 22 miles? And why exactly am I using kilometers?

 

[rl.bhat]

How did you get 5.5 kilometers? Did you use a conversion factor between kilomters and miles from 22 miles? And why exactly am I using kilometers?

Sorry.It is miles.
22 mile*15 min/60 min