# Thread: Velocity and trains..question a was no problem ...

1. ## Velocity and trains..question a was no problem ...

The velocity of a train is given by v = 250 −1/2t^2 .
(a) What is the displacement of thetrain after 5 hours? and 10 hours?

I integrated and got the correct answers for both being
d5 =7375 /6,
d10 =7000 /3
(b) At what times is the displacement of the train zero?

I made 250-1/2t^2 = 0 to find the displacement of the train at 0 but I couldn't work it out...hmm
maybe my thinking is messed up.

(c) Describe what’s happening to the train.

hmmmm....

Thomas the tank engine...

2. ## Re: Velocity and trains..question a was no problem ...

$\displaystyle \int_0^T 250-\dfrac{t^2}{2} \, dt = 0$

$\bigg[250t-\dfrac{t^3}{6}\bigg]_0^T = 0$

$250T - \dfrac{T^3}{6} = 0$

Solve for $T$, the time(s) when the train's displacement is zero.

Do you understand what zero displacement means?

3. ## Re: Velocity and trains..question a was no problem ...

The graph of $\displaystyle v= 250- (1/2)t^2$ is parabola opening downward with vertex at t= 0, v= 250. That means the train starts with its maximum velocity, 250, then slows down. It will stop (velocity 0) when [tex]t= 50\sqrt{2}[tex] and then start going backwards, with negative velocity. It will be back at its starting point, displacement 0, when $\displaystyle 250t- \frac{t^3}{6}= 0$, as skeeter says. We can write that as $\displaystyle 1500t- t^3= t(1500- t^2)$. That has three roots, t= 0, when it started, a negative root, which is irrelevant, and a positive root when it will be back at its starting point.

4. ## Re: Velocity and trains..question a was no problem ...

Thanks Skeeter I never thought of putting T on the b -a part of the integral...I'll have to keep my eye out for that type of situation..i think 0 displacement means the train hasnt moved from its original position right ?....thanks again it really does make me more aware that I can use any value with letters , numbers and symbols for the integral part something I wasn't aware of ie T to 0
Cheers !!

5. ## Re: Velocity and trains..question a was no problem ...

Thanks HallsofIvy The point where its stops comes out at [tex]t= 50\sqrt{2}[tex] so I 'm nt quite sure what that means with the [tex] parts ....but the rest is clear ....where it stops how did you work that out ....thanks

6. ## Re: Velocity and trains..question a was no problem ...

I forgot the "\" in the Latex ending "[\tex]". It should have been $\displaystyle t= 50\sqrt{2}$.

7. ## Re: Velocity and trains..question a was no problem ...

Thanks !! how did you work that out to" stop" the train cheers?

8. ## Re: Velocity and trains..question a was no problem ...

I didn't! I made a foolish error and got the wrong answer! Thanks for pointing that out! $\displaystyle 250- (1/2)t^2= 0$ reduces to $\displaystyle 2(250)= 500= t^2$ so that $\displaystyle t= \pm\sqrt{500}= \pm\sqrt{5(100)}= \pm 10\sqrt{5}$, not $\displaystyle 50\sqrt{2}$! Of course, only the positive answer makes sense in this problem.

9. ## Re: Velocity and trains..question a was no problem ...

No probs , I would've believed it as your math ability is alot higher than mine ...cheers