I recently got asked the problem that follows:

The leader of a bicycle race is traveling with a constant velocity of +14.20 m/s and is 16.6 m ahead of the second-place cyclist. The second-place cyclist has a velocity of +8.70 m/s and an acceleration of +1.15 m/s2. How much time elapses before he catches the leader?

I got the answer 12.89 seconds, although I know this is incorrect. I have no clue how to even begin solving this. Here's what I did as a guess though:

First, I set up my matrix:

Displacement: 16.6

Vi: 8.7

Vf: ?

Acceleration: 1.15

Time: ?

Of the four kinematic equations, the only one that could be used is d = vi*t + 1/2*a*t^2

The problem is I am unable to rearrange this equation to get the desired product, which is time.


2 Answers 2


What if you set up two equations $$x_1 = 14.2t + 16.6$$ and $$x_2 = \frac{1.15}{2}t^2+8.7t$$ and set them equal to each other. Each expressions represents the displacement value of the respective rider and when they equal each other, the 2nd rider has caught the 1st.

  • $\begingroup$ I am trying to algebraically solve the equations by setting them equal to eachother, but I cannot continue any further than "4.925t + 16.6 = t^2" $\endgroup$ Commented Sep 10, 2017 at 19:28
  • $\begingroup$ If you are going to be solving kinematic problems with the formula you outlined in your question, you will most likely need to be able to solve quadratic equations. Have you had any practice solving equations with an $x^2$ (or $t^2$) term in them? $\endgroup$ Commented Sep 10, 2017 at 19:56
  • $\begingroup$ Oh, you're saying I need to use the quadratic equation! Okay I'll give that a shot! $\endgroup$ Commented Sep 10, 2017 at 20:06
  • $\begingroup$ I plugged the equation into the quadratic formula and I got 7.22 as a result, which appears to still be the incorrect answer for time $\endgroup$ Commented Sep 10, 2017 at 20:08
  • $\begingroup$ I think your equation is wrong. After rearranging it I get $.575t^2-5.5t-16.6=0$ with a solution of about $t\approx12$ $\endgroup$ Commented Sep 10, 2017 at 20:14

Simply consider it in terms of a relative frame $s_1-s_2=(s_{0,1}-s_{0,2})+(v_1-v_2)\cdot t+1/2 (a_1-a_2)\cdot t^2)=16.6+5.5 \cdot t-1/2\cdot1.15 \cdot t^2=0$.

The kinematic equation to use actually is $$ s(t)= s(0)+v(0)t+1/2 at^2$$ for constant $a(t)=a(0) = a$.

addendum as per your comment

Sorry for going too much ahead. Suppose you know the equation for $s(t)$ that I put last, which comes from $$ \ddot s(t) = a\quad \Rightarrow \quad \dot s(t) = v(0) + at\quad \Rightarrow \quad s(t) = s(0) + v(0)t + {1 \over 2}at^{\,2} $$ Then just express it for the leader and for the 2nd cyclist, where the starting positions are $s_1(0)=s_2(0)+d$, and find the $t$ value for which $s_1(t)=s_2(t)$

  • 1
    $\begingroup$ I'm sorry. I'm a highschool student in his first few weeks of physics class. I have no idea what that means or how to use it. $\endgroup$ Commented Sep 10, 2017 at 19:52
  • $\begingroup$ pardon. I expanded my answer and hope it is clear now. $\endgroup$
    – G Cab
    Commented Sep 10, 2017 at 21:06

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