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This question is not limited to any particular method; all methods accepted. This is a homework question, stated exactly as:

A cyclist starts off in a bike at an average speed of 16 km/h. 15 minutes later, a motorcyclist sets off on the same trail with an average speed of 48 km/h. How many minutes (in total) will the cyclist have ridden when he gets overtaken by the motorcycle?

I tried to form 2 simultaneous equations for the speed and distance of both but failed. Any help??


EDIT

Thanks for the retag, and you asked to me to show my working out. Knowing that the the first guy crossed d kilos in 15 minutes, we can work that out as 4. Now for the second guy, he would cross those 4 in only 5 minutes. So I tried to plot a linear relationship for each guy and solve for an intersection point, but since they both start at the origin (do they???) The only intersection is (0, 0). And if I try to start off guy#2 at 15 minutes, the intersection point is 30. The answer at the back of the book says 22.5

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    $\begingroup$ I think it would be a good idea for you to show your work and your thoughts. That would be the best way for us to help you. $\endgroup$ – Matthew Conroy Feb 4 '18 at 1:16
  • $\begingroup$ Also better write the equations for the time and distance. $\endgroup$ – user491874 Feb 4 '18 at 1:16
  • $\begingroup$ I'd suggest to read about Zeno's paradox en.wikipedia.org/wiki/… or in here math.stackexchange.com/questions/787125/… $\endgroup$ – Maths Survivor Feb 4 '18 at 1:42
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The speed of the cyclist is 16km/h. You do not know how long it will have been traveling when it is overtaken, so we will just call that time $t$. Because the cyclist has a quarter-hour head start, the time it will have been traveling when it is overtaken will be $t + 0.25$. Distance is speed multiplied by time, and so will be written as $16(t + 0.25)$.

The speed of the motorcyclist is 48/h. Like with the cyclist, the time is unknown. However, since you know that they will pass each other at the same time, the time will be the same, and will therefore be represented by the same $t$. Therefore, the distance the motorcyclist will have traveled will be written as 48t.

You want to find the time when they pass each other, aka when their distances are equal, so put it all in a simultaneous equation. $$16(t+0.25) = 48t$$

This simplifies to $t = 0.125$. Add on the quarter hour head start to get 22.5 minutes.

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  • $\begingroup$ Your statement "$t$ is the time in hours" is vague. Can you see how to clarify this? $\endgroup$ – Matthew Conroy Feb 4 '18 at 1:32
  • $\begingroup$ @MatthewConroy Is this better? $\endgroup$ – Leif Metcalf Feb 4 '18 at 1:45
  • $\begingroup$ I see, same method but different perspective than Linda's answer. $\endgroup$ – TGamer Feb 4 '18 at 3:29
  • $\begingroup$ @LeifMetcalf Yes, that is a nice improvement. Cheers! $\endgroup$ – Matthew Conroy Feb 4 '18 at 7:32
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The speed of the cyclist is $3$ times slower than the speed of the motorcyclist. The motorcyclist, however goes after $15$ minutes from the same point. You can use number sense to figure this.

Or, when $x=$ amount in hours,

$16x=48(x-0.25)$, since the cyclist sets in $15$ minutes before the motorcyclist.

$0.25$ hours = $15$ minutes

$16x = 48x-12$

$-32x=-12$

$x=\cfrac {3}{8}$

The motorcyclist will overcome the cyclist after $\cfrac {3}{8}$ hours, or $22$ minutes and $30$ seconds.

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    $\begingroup$ It is a 15 minute head start, not a 15 kilometre head start. $\endgroup$ – Leif Metcalf Feb 4 '18 at 1:32
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enter image description hereThe speed of the cyclist is v1=16km/h and the speed of motorcyclist is v2=48km/h, when they two will meet the distance passed by each of them will be equal so d1=d2 but the time that took the motorcyclist to pass that distance is 15 minutes less than the time needed for cyclist to pass the same distance, let's write t2=t1-15.

Therefore we have this system of equations:

d1=v1*t1

d2=v2*t2

From d2=d1 and t2=t1-15 we find that:

v1*t1=v2*(t1-15)

Replacing their values we'll have:

16km/h*t1=48km/h(t1-15)

Solving this we get:

t1=45/2 => t1=22.5 or 22 minutes and 30 seconds

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    $\begingroup$ Please use mathjax in your answers. Thank you! $\endgroup$ – Gaurang Tandon Feb 4 '18 at 2:00
  • $\begingroup$ Well I guess everyone understands what I wrote so there's no need to use mathjax. The answer is pretty clear. $\endgroup$ – Maths Survivor Feb 4 '18 at 2:09
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    $\begingroup$ I think it's best to just use MathJax in most places, even for trivial math, since it is the most specific and clear way of rendering math that we have. $\endgroup$ – Jaideep Khare Feb 4 '18 at 2:34
  • $\begingroup$ Just for clarification, how can I graphically show that? $\endgroup$ – TGamer Feb 4 '18 at 3:38
  • $\begingroup$ @TGamer Does this photo help? $\endgroup$ – Maths Survivor Feb 4 '18 at 5:17
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The time will be infinite, because:

  • When the motorcyclist sets off, the cyclist will be at some point $A_1$ down the road.
  • By the time the motorcyclist reaches the point $A_1$, the cyclist will be at some point $A_2$ further down the road.
  • By the time the motorcyclist reaches the point $A_2$, the cyclist will be at some point $A_3$ further down the road,

and so on. This can be argued to infinity, thus the motorcyclist can never overtake the cyclist.

Edit: Of course, this was a spoof answer. However, when the motorcyclist starts, as they are $3$ times faster than the cyclist, it will take them $15/3=5$ min to make it to $A_1$, then $5/3$ min to make it to $A_2$ etc. - so the total time to overtake the cyclist is $5+\frac{5}{3}+\frac{5}{3^2}+\cdots=\frac{5}{1-\frac{1}{3}}=\frac{15}{2}=7.5$ minutes (the sum of a geometric progression). Add to that the initial $15$ min, and you get that the cyclist was riding for $22.5$ min before they were overtaken.

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  • $\begingroup$ Couldn't resist... $\endgroup$ – user491874 Feb 4 '18 at 1:20
  • $\begingroup$ And yet I get passed all the time! $\endgroup$ – Matthew Conroy Feb 4 '18 at 1:28
  • $\begingroup$ @user873617 is this somehow related to Zeno's Paradox or maybe I'm wrong ? $\endgroup$ – Maths Survivor Feb 4 '18 at 1:28
  • $\begingroup$ @Linda, Yes, this answer was supposed to be a little homage to it, until the OP or someone else comes up with a proper answer. en.wikipedia.org/wiki/… $\endgroup$ – user491874 Feb 4 '18 at 1:30
  • $\begingroup$ @Linda this doesn't make sense. The second guy is faster. It is like we race, and I am slower than you but you start after me you will overtake me. $\endgroup$ – TGamer Feb 4 '18 at 3:17

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