Finding time from average speed

[a lone fishy]

Homework Statement

A golfer rides in a golf cart at an average speed of 3.1 m/s for 28s. She then gets out of the cart and starts walking at an average speed of 1.3 m/s. For how long must she walk if her average speed for the entire trip, riding and walking, is 1.8 m/s ?

Homework Equations

[/B]
Vave = distance/time

The Attempt at a Solution

[/B]
This question seems simple considering there's only formula at play. I tried splitting up into 2 parts. The first part, the person walks at an average speed of 3.1 m/s for 28s. Distance can be calculated as 86.8 m

After this I got lost. I know there's a second part where she walks at an average speed of 1.3 m/s. However nothing else is known. And we know her entire average speed is 1.8 m/s

Any ideas?

 

[Samy_A]

Let $t$ be the time she walks.
You can compute the total time of the trip and the total distance of the trip in function of $t$.
As you also know the average speed for the entire trip, this will give you an equation in $t$.

 

[a lone fishy]

I didn't quite understand your second line.
Right now I have 1.8=86.8 + d / 28 + t where 1.8 is the average speed of the whole trip, d is the unknown distance from the second interval, and t is the unknown time from the second interval.

Im not sure if I am on the right track nor where to proceed from here

EDIT
where does the 1.3 m/s speed in the second interval come into play?

 

[Samy_A]

I didn't quite understand your second line.
Right now I have 1.8=86.8 + d / 28 + t where 1.8 is the average speed of the whole trip, d is the unknown distance from the second interval, and t is the unknown time from the second interval.

Im not sure if I am on the right track nor where to proceed from here

I assume you meant $1.8=(86.8 + d) /( 28 + t)$.

You know that she walked at 1.3 m/s, so that should allow you to express $d$ in terms of $t$.

 

[jbriggs444]

Minor correction: she walks at 1.3 m/s

 

[Samy_A]

Minor correction: she walks at 1.3 m/s

Thanks, I corrected the typo.

 

[Samy_A]

EDIT
where does the 1.3 m/s speed in the second interval come into play?

That's the correct question. You haven't used the 1.3 m/s speed so far.
What relation is there between 1.3 m/s, d and t?

 

[a lone fishy]

Ah I think i got it now.

Right now we have the following: 1.8=(86.8+d)/(28+t).

If we find that d then we can have an equation solvable for t.

In interval two we have 1.3 = distance / t
and this equals to: distance = 1.3t

Then we can plug in that value as the d and solve for t :D