Simple kinematics question -- Riding a bicycle race in two different gears

[knapklara]

Homework Statement

A cyclist does 42 km long race. Because of his shifting preferences he races the race with v1= 10 m/s and v2=11.5 m/s. What distance does he do with v1 and what with v2? The whole race lasted 64 minutes. I know the basic formulas but somehow I am lost. Help is much much appreciated!

Relevant Equations

s = v x t
t = t1 + t2
s1 = v1 X t1

I calculated average velocity but obviously it helps nothing with this problem.
I hope to get me going with these exercises once I break the ice. Thank you in advance!

 

[Lnewqban]

Welcome!
Could you post your work so far?

 

[kuruman]

You skipped a relevant equation. You have
t = t1 + t2
s1 = v1*t1

That's a good start. Do you see what other equation is missing?

 

[knapklara]

I'm really not sure what I'm missing, that's the problem :( I'm a little rusty. Don't know how to proceed at all.

 

[knapklara]

I thought about getting t2 from t1, but that still doesn't help one bit. I know the answer is right in front of me, but I just don't see it.

 

[Lnewqban]

If you draw a graph of velocity versus time, the average velocity that you have calculated would be represented by a horizontal line that runs between times 0 and 64 seconds.
The area of that rectangle would represent the total distance of 42,000 meters.

Using the two given velocities, you should end up with two rectangles in that graph, which total area should also represent 42,000 meters.
Adjusting the point in time at which the cyclist change velocities, you should find the areas (and therefore, covered distances) corresponding to V1 and V2.

 

[kuruman]

I thought about getting t2 from t1, but that still doesn't help one bit. I know the answer is right in front of me, but I just don't see it.

You are missing the equation for s2 (it's in your notes but not your post #1) and two simple facts: t2 = 64 min - t1 and s2 = 42 km - s1. Or you can do it geometrically as @Lnewqban suggested.

 

[knapklara]

You are missing the equation for s2 (it's in your notes but not your post #1) and two simple facts: t2 = 64 min - t1 and s2 = 42 km - s1. Or you can do it geometrically as @Lnewqban suggested.

Allright, that makes more sense, but I still don't know what to do with two unknown variables. Please forgive my ignorance.

 

[kuruman]

You have
s1 = v1 t1
and
s2 = v2 t2 which becomes
42 km - s1 = v2 (64 min - t1)
What do you get when you substitute s1 from the top expression?

When you put in the rest of the numbers, don't forget to convert kilometers to meters and minutes to seconds.

 

[knapklara]

All right, I'd kindly ask you for a comment. I knew all along that it was going to be quite simple and thank you for bearing with me! You pointed the obvious!

 

[kuruman]

All right, I'd kindly ask you for a comment. I knew all along that it was going to be quite simple and thank you for bearing with me! You pointed the obvious!

Not to worry. I too have the occasional blind spot and can't see the obvious.

 

[haruspex]

All right, I'd kindly ask you for a comment. I knew all along that it was going to be quite simple and thank you for bearing with me! You pointed the obvious!

That's all fine, but you can get there more quickly if you keep your eye on what needs to be done at each step.
We start with the facts $v_1t_1+v_2t_2=s$, $t_1+t_2=t$.
We have two equations and two unknowns. So we pick one unknown and use one equation to eliminate that unknown from the other:
$t_2=t-t_1$
$v_1t_1+v_2(t-t_1)=s$
Our remaining unknown occurs more than once, so collect the terms involving it together:
$(v_1-v_2)t_1+v_2t=s$
Yes, you did all that, but with several unnecessary digressions along the way.

This process extends to any (solvable) system of n equations and n unknowns.

 

[knapklara]

Thank you very much for your perspective, much appreciated!