Lap time physics/logic problem

  • Thread starter nokia8650
  • Start date
  • #1
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I was struggling with the question below:

Two drivers take part in a race. The faster driver’s average lap time is 1 minute 6 seconds,
and the slower driver’s 1 minute 10 seconds.

Assuming they drive at a constant speed and started together, how long would it take for
the faster to lap the slower driver?

A 18 minutes 9 seconds

B 19 minutes 15 seconds

C 20 minutes 25 seconds

D 38 minutes 30 seconds

E 40 minutes 50 seconds


The correct answer is B - and I can see how this is arrived at, however I do not know why this is correct, as opposed to A:

My working:

4 sec difference in lap times

66 seconds/4 = 16.5 laps

16.5*66 = 18 minutes 9 seconds


Clearly, for the correct answer, rather than 66 seconds/4 = 16.5 laps, you must use 70 seconds (the slower driver's time). Can someone please explain the logic behind this?

Thanks
 

Answers and Replies

  • #2
Questions like this are really abstract, because you don't know some information that is seemly obvious to know (like the length of the lap). So why don't you assume one like "x_lap"? In that case, the speed of the fast car is x_lap/66. The speed of the second car is x_lap/70. When the cars overlap at time t, the distance for the slow car is x_final, the distance for the fastcar is x_final + x_lap.

Your equations of motion for the positions of the cars at the final time t (YOU fill in):
Slow car (equation 1) x_final =
Fast car (equation 2): x_final +x_lap =

You want to find t, and boy doesn't this look horrid because you have 3 variables and 3 unknowns (x_lap, x_final, and t)! But it really isn't bad. Subtract the first equation from the second, and you SHOULD find:

x_lap = x_lap/66*t - x_lap/70*t

Now divide this by x_lap (it's in everything! ... and you should always eventually look for things you make assumptions about to drop out!):

1 = 1/66*t - 1/70*t = (1/66 + 1/70)*t

So t = ?

Bingo!

That's MUCH more straight-forward than trying to do something abstract, like look at differences between velocities and abstract this to some future condition, as you originally tried to do.

P.S. I worked this all out using an unknown lap distance, x_lap. I often suggest to my students that they COULD (in a pinch like a test) use something simpler, like 1 mi... IF it really seems in the problem they could do so. But you DO have to be careful about this technique... it's really better to leave assumed conditions as variables to be sure they cancel out.
 
Last edited:
  • #3
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Your working seems to be just playing with numbers. Why would you divide the period of the faster one by the difference? Can you explain your rationale?


The straightforward way to put it (at least for me) is
d1=d2+2Pi*r
The fast one has one lap advance when he first passes the slow one.
or even better, in terms of angles,
theta1=theta 2 + 2*Pi

Then you can manipulate this in various ways, by using speeds or periods.
For example,
theta1= (2*pi/T1)*t
theta2= (2*Pi/T2)*t
T1,T2 are the periods, given in the problem
 

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