# Finding the least acceleration required for particle $A$ to overtake particle $B$

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.

• What is the initial velocity of A? – Paul Mar 16 '18 at 20:53
• Same as B, at 5 m/s – Benny Mar 16 '18 at 20:54

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 $a$.

• 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 – Benny Mar 16 '18 at 21:00
• Check your arithmetic. I get 0.2 from this equation. – Paul Mar 16 '18 at 21:03
• Oh I see what I did wrong, thanks a bunch – Benny Mar 16 '18 at 21:08

hint

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

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

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