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You have two runners modeled as particles $A$ and $B$ respectively. Say that particle $B$ is ahead of particle A by $10$ m, and particle $B$ has a constant velocity of $5$ m/s. When particle $B$ is $50$ m from the finish line particle $A$ starts to accelerate, but $B$ does not.

My question is, what is the least acceleration $A$ must produce in order to overtake $B$?

So far, I have tried the following:

  • Knowing that particle $B$ is traveling at a constant velocity, I have calculated the time it would take for particle $B$ to cross the finish line, by using $v= \dfrac st$. The time I got was $10$ s.
  • Since particle $A$ is $10$ m away from particle $B$, and since I know that particle $A$ is $2$ s behind particle $B$ $\left(\text{by using }v= \dfrac st\right)$, I used the equation of kinematics to work out the acceleration as I knew the distance, the time, and the initial speed.

However, I cannot seem to get the correct answer.

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  • $\begingroup$ What is the initial velocity of A? $\endgroup$ – Paul Mar 16 '18 at 20:53
  • $\begingroup$ Same as B, at 5 m/s $\endgroup$ – Benny Mar 16 '18 at 20:54
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A has 60 meters to run and has to make it in 10 seconds, with an initial speed of 5 m/s. The constant acceleration equation is $$d=v_0 t +1/2 at^2$$ And you just need to solve for $a$.

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  • $\begingroup$ That is what i have done and the answer that I got was 0.8 m/s^2, however, the correct answer is 0.2 m/s^2 $\endgroup$ – Benny Mar 16 '18 at 21:00
  • $\begingroup$ Check your arithmetic. I get 0.2 from this equation. $\endgroup$ – Paul Mar 16 '18 at 21:03
  • $\begingroup$ Oh I see what I did wrong, thanks a bunch $\endgroup$ – Benny Mar 16 '18 at 21:08
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hint

A has to cross $d=10+50 m $ in less than $10s $. with

$$d=\frac {1}{2}at^2+5t$$ solve the inequation $t <10$

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  • $\begingroup$ Your question is not well asked. $\endgroup$ – hamam_Abdallah Mar 16 '18 at 21:02

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