A body of mass $m$ has initial velocity $v_0$ in the positive $x$-direction. It is acted on by a constant force $F$ for time $t$ until the velocity becomes zero; the force continues to act on the body until its velocity becomes $−v_0$ in the same amount of time. Write an expression for the total distance the body travels in terms of the variables indicated.

I let $t'$ denote the time at which the velocity goes to zero so I can use $t$ as a variable. Then I found

$$a = \frac{\Delta v}{\Delta t} = \frac{-2v_0}{2t'} = -\frac{v_0}{t'}$$

Substituting this expression for $a$ into $s(t) = \frac{1}{2} a t^2 + v_0 t$ yields

$$\begin{align} s(t) & = \left(-\frac{1}{2} \frac{v_0}{t'}\right) t^2 + v_0 t \\ s(2t') & = \left(-\frac{1}{2}\frac{v_0}{t'}\right)(2t')^2 + v_0 (2t')^2 = 0 \end{align}$$

This makes intuitive sense to me. If the positive $x$ direction is up and the force is gravity, then if we shoot a ball upward with velocity $v_0$, it will decelerate to velocity $-v_0$ exactly when it returns to the launch position.

To double check, I tried taking (from $a = -\frac{v_0}{t'}$ and $F = ma$) $a = \frac{F}{m}$ and $v_0 = -\frac{F}{m}t'$. I get the same answer when substituting into $s(t)$:

$$\begin{align} s(t) & = \frac{1}{2} a t^2 + v_0 t \\ & = \left(\frac{1}{2}\frac{F}{m}\right) t^2 -\left(\frac{F}{m}t'\right) t \\ \implies s(2t') &= \left(\frac{1}{2}\frac{F}{m}\right) (2t')^2 -\left(\frac{F}{m}t'\right) (2t') =0 \end{align}$$

My textbook, however, says without comment that the answer is $\frac{F}{m}(t)^2$, or in my notation, $\frac{F}{m}(t')^2$. Am I wrong, or is the textbook wrong, or could this be an ambiguity in wording—e.g. is there a way to construe "total distance the body travels" as "the furthest position the body reaches from its starting point"? Or could it be that the question is asking not about the object's location at time $t = 2t'$ but about its location as a function of time $s(t)$? In that case my answer is found above and would still be wrong.

  • $\begingroup$ I think it comes down to this. If you walk 100m in the x direction and then walk directly back to where you started from, is the distance yoou have travelled 200m or 0m? $\endgroup$
    – Dr Chuck
    Apr 30, 2020 at 8:03
  • $\begingroup$ I could argue for either, but the textbook answer looks to me like the equivalent of 50m! $\endgroup$
    – Max
    Apr 30, 2020 at 8:09
  • $\begingroup$ Why not use $v^2=v_0^2+2as$ instead? $\endgroup$
    – bemjanim
    Apr 30, 2020 at 8:11
  • $\begingroup$ @bemjanim Then $s(t) = \frac{v_0^2 - v(t)^2}{2 F/m}$, and at $t = 2t'$, $v(t) = -v_0$ and $s(2t')=0$, as above. $\endgroup$
    – Max
    Apr 30, 2020 at 8:41
  • $\begingroup$ Your quantity $s(t)$ is the coordinate of the object at time $t$, i.e. the distance from the origin of the implicit reference frame. It should not be confused with the distance travelled. $\endgroup$ Apr 30, 2020 at 8:50

2 Answers 2


The total distance is a sum of two distances: $S_{1}$ and $S_{2}$, where the first is till the velocity goes to zero, the second: from zero to $-V_{0}$. $S_{1}=V_{0}t+\frac{1}{2}at^{2}$. Here $a=\frac{-V_{0}}{t}$. So $S_{1}=V_{0}t-\frac{V_{0}t}{2}=\frac{V_{0}t}{2}$. The same for $S_{2}$ but with zero initial speed and positive acceleration: $S_{2}=\frac{1}{2}at^{2}=\frac{1}{2}\frac{V_{0}}{t}t^{2}=\frac{V_{0}t}{2}$

Total $S=S_{1}+S_{2}=V_{0}t=\frac{F}{m}t*t=\frac{F}{m}t^{2}$.

  • $\begingroup$ You write $a=−V_0 /t$, but the $t$ in the denominator here refers to the specific time (that I denote $t'$) at which the velocity reaches zero, not the time variable $t$. Therefore, I don't think you can cancel it out as you do when you change $at^2 /2$ to $V_0 t /2$. $\endgroup$
    – Max
    Apr 30, 2020 at 8:31
  • $\begingroup$ To put it more bluntly, if $a(t) = -v_0 / t$, then at $t=0$ the acceleration is undefined! $\endgroup$
    – Max
    Apr 30, 2020 at 8:37
  • 1
    $\begingroup$ You overcomplicated the problem very much. There is no need for any $t_{'}$, you just have two distances of uniformly accelerated motion, both with time duration of $t$. About your last comment, you misunderstand the definition of acceleration: it is just the rate at which velocity is changed. In other words, the change of the velocity from $V_{0}$ to $0$ divided by the change in time which is $t$ of course. Sure, the acceleration is undefined, since there is no motion when $t=0$, so no change in time. $\endgroup$ Apr 30, 2020 at 8:43
  • $\begingroup$ OK, I get it now, although I think that rather than saying acceleration is undefined at $t=0$, you mean to say that the acceleration is constant throughout the problem. The reason I am attached to the distinction between $t$ and $t'$ is that I want to have a function $s(t)$ that's valid for all $t$. Using your equations, it's hard to produce an expression for $s(1.3 t')$, for instance, because it's not clear which $t$ you should substitute. $\endgroup$
    – Max
    Apr 30, 2020 at 9:13

The textbook's answer reflects the total distance traveled, i.e.

$$\vert s(t') - s(0) \vert + \vert s(2t') - s(t') \vert = \vert \frac{1}{2} v_0 t' - 0 \vert + \vert 0 - \frac{1}{2} v_0 t' \vert = v_0 t' = -\frac{\vec F}{m}(t')^2 = \frac{F}{m}(t')^2$$

I prefer this way of organizing it because it lets me treat the displacement as a single function $s(t)$ and preserves the distinction between a specific value $t'$ and the variable $t$ that I am accustomed to in mathematics. Others may find the answer from @Anton Baranikov more legible.


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