1. ## velocity help

I solved the problem but I want to make sure I did it correctly.

starting at time t=0, a train is halted by a braking deceleration of $\frac{1}{2}\frac{m}{s^2}$, with the result that its position at time t is $s(t) = s_{0} +v_{0} -\frac{1}{4}(t^2)$

show that $s_{0}$ is the train's postion when the brakes are applied and $v_{0}$ is its velocity at that time. Find the stopping distances if $v_{0}$ is $20 \frac {m}{s}$

$s_{0} +v_{0}*(0) -\frac{1}{4}(0^2) = 0$ gives $s_{0} =0$

and

$v_{0} -\frac{1}{2}(0) = 0$ gives $v_{0} =0$

and solve s(t) to get the distance using $v_{0} = 20$

$20 -\frac{1}{2}t\Rightarrow t =40$

$s(t) = 0 + 20(40) -\frac {1}{4}(40^2) = 400$

thus the answer is 400 metres.

ps. teaching myself LaTex

2. ## Re: velocity help

Originally Posted by Jonroberts74
I solved the problem but I want to make sure I did it correctly.

starting at time t=0, a train is halted by a braking deceleration of $\frac{1}{2}\frac{m}{s^2}$, with the result that its position at time t is $s(t) = s_{0} +v_{0} -\frac{1}{4}(t^2)$

show that $s_{0}$ is the train's postion when the brakes are applied and $v_{0}$ is its velocity at that time. Find the stopping distances if $v_{0}$ is $20 \frac {m}{s}$

$s_{0} +v_{0}*(0) -\frac{1}{4}(0^2) = 0$ gives $s_{0} =0$

and

$v_{0} -\frac{1}{2}(0) = 0$ gives $v_{0} =0$

and solve s(t) to get the distance using $v_{0} = 20$

$20 -\frac{1}{2}t\Rightarrow t =40$

$s(t) = 0 + 20(40) -\frac {1}{4}(40^2) = 400$

thus the answer is 400 metres.

ps. teaching myself LaTex
Stopping distance is s(t) - s0 = d = v0 - (a/2)t^2

v0 is not zero, it is given 20 m/s

also you need to know v(t) - v0 = - a*t what is v(t)? it is given in the statement of the problem.
Does this help

3. ## Re: velocity help

okay so

$s(t) = s_{0} + v_{0} -\frac{1}{4}(t^2) \Rightarrow s(t) - s_{0} = d = v_{0} - \frac{1}{4}(t^2)$

and $v(t) - v_{0} = - a(t)$ ??

4. ## Re: velocity help

Originally Posted by Jonroberts74
ps. teaching myself LaTex
Nice job on the LaTeX.

Did you copy the problem word for word? Some of this just doesn't make sense.
Originally Posted by Jonroberts74
starting at time t=0, a train is halted by a braking deceleration of $\frac{1}{2}\frac{m}{s^2}$, with the result that its position at time t is $s(t) = s_{0} +v_{0}t -\frac{1}{4}(t^2)$

show that $s_{0}$ is the train's postion when the brakes are applied and $v_{0}$ is its velocity at that time. Find the stopping distances if $v_{0}$ is $20 \frac {m}{s}$
I am very confused about the first two questions. You didn't get them right, at least the way the question is worded. $s_0 = 0$ can be defined at the starting point, but that was never mentioned, and $v_0$ had better not be 0 else the train isn't moving at t = 0.

The brakes are applied at t = 0. I'm not sure what they wanted out of the first question...we need to show that the position of the train at t = 0 ( $s_0$) is the train's position. So we have
$s(0) = s_0 + v_0(0) - \frac{1}{4}(0)^2 = s_0$ So $s(0) = s_0$ is the position of the train when t = 0.
(The acceleration is opposite to the speed of the train. Good job in picking up the negative there.)
And v_0 is equal to v at t = 0. Well, we have
$\frac{ds}{dt} = v(t) = v_0 - \frac{1}{2}t$

at t = 0:
$v(0) = v_0 - \frac{1}{2}(0) = v_0$ So $v(0) = v_0$ is the speed of the train at t = 0.

Why were these questions even asked?? $s(0) = s_0$ and $v(0) = v_0$ are from the definitions of s(0) and v(0). Again we can usually assume $s_0 = 0$, but this was never done. I just don't get it.

For the last part we know that v(t) = 0 at the final point, so from the speed equation we have
$0 = 20 - \frac{1}{2}t \implies t = 40$, as you said, but:

$s(40) = s_0 + 20(40) - \frac{1}{2}(40)^2$

What is $s_0$? This was not given in the problem statement. (Typically we set $s_0 = 0$ by choice of the origin, but this was never mentioned for this problem.)

Tell your professor for me that you need to set an origin for Physics problems (which is why I'm repeating that.) If you take the train to be at the origin then we have $s(0) = s_0 = 0$ then the last question can be done.

-Dan

Edit: We cross posted. Using $s(40) - s_0 = d$ is a good move. Thanks votan.

5. ## Re: velocity help

and

$v(t) = 20 - \frac{1}{2}(t), t=0 \Rightarrow v(t) = 20$

6. ## Re: velocity help

Originally Posted by Jonroberts74
and

$v(t) = 20 - \frac{1}{2}(t), t=0 \Rightarrow v(t) = 20$
You had that right in the first post. The train has stopped at some time t. You are trying to find t, so 0 = 20 - (1/2)t, etc.

-Dan

7. ## Re: velocity help

yeah that is word for word the problem.
yes it is very vague, I had assumed it was referring to s_0 as being the origin.

thank you!

8. ## Re: velocity help

Originally Posted by Jonroberts74
yeah that is word for word the problem.
yes it is very vague, I had assumed it was referring to s_0 as being the origin.

thank you!
Perhaps I was a bit manic over that. But I have a real pet peeve when most Calculus teachers try to teach kinematics problems. Their terminology is typically awful, as are their assumptions...they rarely define the origin, much less define positive directions. They simply assume one and never say.

-Dan

9. ## Re: velocity help

Yes the statement of the problem is poor and confusing to most students. I picked up what the question was about from the statement - find the stopping distance. I don't know if this this a textbook question, or the instructor lousy excess of zeal to complicate the life of his students. He could be a lab technician in charge of the course, and that shows how much he has to learn before leading a class on this subject.

10. ## Re: velocity help

Unfortunately, I do not have a professor. I am teaching myself math in the hopes to get into university to study the maths. It can be frustrating but I am persistent. I do see the error in the textbook. I really like differential geometry, I read about it but still a lot of it is beyond my current knowledge base--I will get there.