Answer: $a=v^2/d$ is the formula I needed.

This is a problem in a programming assignment, I haven't taken physics.

I have a starting speed (0 m/s), an final speed (208.33m/s) and the distance it took to reach that speed (200km). I need to get the time it will take to travel any arbitrary distance lower than 200km.

From what I remember in highschool, I used basic calculus to get $a/2(t^2)=distance$ or for what I have $t=\sqrt(400000(m/s)/a)$. what I don't know is the acceleration, usually to calculate acceleration I'd need time. I'm not sure where to go from here.


closed as off-topic by Abhimanyu Pallavi Sudhir, John Rennie, tpg2114, Michael Brown, Qmechanic Nov 2 '13 at 12:12

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  • $\begingroup$ Is the acceleration constant throughout the time interval in which the speed increases? $\endgroup$ – GTX OC Oct 4 '13 at 5:08
  • $\begingroup$ Yes. I already have the answer though. Thanks for suggesting the edit. $\endgroup$ – Dan Oct 4 '13 at 5:09
  • $\begingroup$ The "formula I needed" is not correct... $\endgroup$ – DJohnM Oct 4 '13 at 6:29
  • $\begingroup$ Oh well, the assignment was due before your post. It's not a big grading factor, I'm not sure anyone got that part right. $\endgroup$ – Dan Oct 7 '13 at 5:49

Converting all units to SI: $$u=0$$ $$v = 208.33 \text{ m/s}$$ $$d =200000\text{ m}$$So: $$a=\frac{v^2-u^2}{2d}=0.1085\text{ m/s}^2$$

Then, for any distance,$d$ in meters:$$t=\sqrt{\frac{2d}{a}}$$This vehicle maintains a feeble, but constant, $0.01\text{ g}$ for a $200\text{ km}$ trip lasting a little more than 30 minutes.


According to the question while the car travels from 0 to 200 m/s with constant acceleration it covers a total distance of 200km.

Applying the formula
where v is the final speed,u is the initial speed and s is the distance. Here u=0m/s,v=208.3m/s and s=200km.
Putting the values we get
$$a=0.1085034 m/s$$

Finally using $$s=ut+1/2at^2$$ where s is the distance covered, u is the initial velocity,a is the acceleration and t is the time, we get the distance(in m) covered at any time t(in seconds).

Here u=0m/s. We get $s=1/2at^2$. Given any s<200km you can now find the time it takes to cover that distance.

  • $\begingroup$ Wouldn't it be 0.1085 since 2*200km is at the bottom, not 2*200m? $\endgroup$ – Dan Oct 4 '13 at 5:42
  • $\begingroup$ This answer seems to use $\frac{km}{hr^2}$ as its unit for acceleration... $\endgroup$ – DJohnM Oct 4 '13 at 6:07
  • $\begingroup$ @Dan: yeah you are correct. I took the speeds in km/h. $\endgroup$ – GTX OC Oct 4 '13 at 6:21

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