Is it possible to calculate the shortest time it takes a body to travel a certain distance if the only information provided is its maximum acceleration, its maximum retardation and the distance it travels (from rest to rest)? I've been trying to work it out but I was wondering if it is necessary to know either its maximum velocity or the distance at which the body switches from accelerating to decelerating.

The maximum acceleration of a body is $9ms^{-2}$ and its maximum retardation is $3ms^{-2}$.
What is the shortest time in which the body can travel a distance of 7200 m from rest to rest?

Answer: $80m$

I have no idea how this answer was reached, I've made velocity-time graphs and tried simulations equations etc. but I can't seem to reach this answer, best I could do was, if the object kept accelerating at $9ms^{-2}$ it would travel 7200 m in 40s. $7200=0t+\frac129t^2$.

This is my first question so I hope I asked it correctly, any help would be greatly appreciated as it is hurting my brain :/

  • 3
    $\begingroup$ Are you in a calculus-level physics course? You don't need calculus, just checking. Hint: break the problem into the 'speeding up' distance and the 'slowing down' distance, each of which obeys $ S = V_0 + at^2/2$ , and make the sum of the two distances equal to the total travel distance. $\endgroup$ – Carl Witthoft Apr 16 '14 at 11:36

The velocity after a time t1 of accelerating is the starting velocity of the deceleration phase. Thus $$a t_1 = b t_2$$ (not worrying about the sign here. I suppose you could)

Further you have $$\frac12at_1^2+\frac12bt_2^2=s$$

Now you have two equations with two unknowns.

Solving: rearrange first equation $$ \frac{a}{b} = \frac{t_2}{t_1}\\ t_2=\frac ab t_1\\ t_2^2=\left(\frac ab\right)^2 t_1^2 $$ Now substitute into second equation: $$ \frac12at_1^2+\frac12\frac{a^2}{b}t_1^2=s\\ $$

Substitute values: $$9t_1^2 + 27t_1^2=14400\\ t_1^2=\frac{14400}{36}\\ t_1=20\\ t_2=3t_1\\ t_1+t_2=80$$

  • $\begingroup$ a_{t} or a_t $\implies a_{t}$ $\endgroup$ – jinawee Apr 16 '14 at 11:54
  • $\begingroup$ I use $[random stuff]$. You can check meta.math.stackexchange.com/questions/5020/… $\endgroup$ – jinawee Apr 16 '14 at 12:07
  • $\begingroup$ I worked out t_1 and t_2 and added them together and got 77.27, did I do something wrong or did the book just round the answer up to 80? Thanks for your help :D $\endgroup$ – Faust Apr 16 '14 at 12:11
  • $\begingroup$ @Faust - you must have made a mistake. See fully worked solution. These are "nice round numbers". $\endgroup$ – Floris Apr 16 '14 at 12:12
  • $\begingroup$ @jinawee - thanks for the hint. Google was not helping me. Thank goodness for humans. You are human, right? :-) $\endgroup$ – Floris Apr 16 '14 at 12:14

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