Math problem involving reciprocal of linear functions

[Matt1234]

Homework Statement

http://img97.imageshack.us/img97/1521/lastscano.jpg

Homework Equations

avg speed = distance / time

The Attempt at a Solution

not sure how to go about this one, looking for a few hints on how to get the equation.

answer is :
t = 3850 /v

But I am not sure why.

thank you

 

[Mark44]

Homework Statement

http://img97.imageshack.us/img97/1521/lastscano.jpg

Homework Equations

avg speed = distance / time

You show the equation above as being relevant, but I'm not sure that you are really convinced. Can you solve this equation for time in terms of the other two variables.

The Attempt at a Solution

not sure how to go about this one, looking for a few hints on how to get the equation.

answer is :
t = 3850 /v

But I am not sure why.

thank you

I'm not sure why either. There's either a typo in your answer sheet or you wrote it incorrectly.

 

[Matt1234]

just confirmed the answer it is as i typed it.

 

[Anakin_k]

t = distance / v

distance = t * v
distance = 11 * 350
distance = 3850

We know distance is 3850km since it takes 11 hours to reach from Quebec City to Montreal at a speed of 350km/h. Time is also inversely proportional to speed so that means as time goes up, speed goes down and vice-versa.

t = 3850/v

To conform this is correct, we will plug the original speed into the equation.
t = 3850/350
t = 11h

If time was lower than 11 (we'll use 5h), speed should go up.
t = 3850/v
5 = 3850/v
v = 3850/5
v = 770 km/h

If time was higher, speed should go down. So let's use 20 hours.
t = 3850/v
20 = 3850/v
v = 192.5 km/h

 

[Matt1234]

thank you very much, that helped a lot. Pretty easy i just didnt see it. :)

 

[Anakin_k]

Anytime! Good luck. :)