# Homework Help: Dynamics question

1. Nov 27, 2008

### Melawrghk

1. The problem statement, all variables and given/known data
A car's velocity can be described by v=6-0.3s. Find its acceleration at t=100s

2. Relevant equations
v=ds/dt, a=dv/dt

3. The attempt at a solution
This was actually a two part question, where the first part required you to find s at t=100.
So I wrote:
ds/dt = 6-0.3s and integrated that equation to get:
s=(e-0.3t-6)/(-0.3)
From which I found that at t=100s, s=20m

For the second part,
a=dv/dt=(dv/ds)*v
To find dv/ds term, I took the derivative of v=6-0.3s and got v'=-0.3. I put the known formulas into the equation above to get:
a=-0.3(6-0.3s)

I know that at t=100s s=20, so I can find the acceleration at t=100 (equivalent to s=20):
a=-0.3(6-0.3*20)=0

I think my thought process makes sense (it usually does), but I might be doing something stupid and not realizing it. Long story short, I just don't like a=0. However, I know some people got 0 as well and some actually got a number (this was on a midterm).

Is it right? And if not, can you please point out the area where I'm doing something wrong?

2. Nov 27, 2008

### naresh

What is your initial condition on s? Let's assume that s=0 at t = 0 (a reasonable assumption). Your expression s=(e-0.3t-6)/(-0.3) does not satisfy this. So you might want to check your integration.

As you say, your thought process makes sense

Edit: You also need to be careful about rounding off numbers too soon. You are effectively saying e-0.3t = 0 at t = 100. While this is okay in the first part, it might not be okay in the second part!

Last edited: Nov 27, 2008
3. Nov 27, 2008

### Melawrghk

So the rounding thing isn't really something I can change. We're only allowed to use simple solar power calculators for this and whenever I put e-30-6 it just gave me -6. The TI-83 does that too.

As for the integration, here is the process:

Maybe my mistake is there? Although this is like the fourth time I've done it today and I still get the same thing. It could be something fundamental I guess.

Soo.. I don't know. Any more ideas?

4. Nov 27, 2008

### naresh

The integral itself is correct, what is wrong is in the application of the limits on the LHS. What do you get on the LHS when s = 0 (lower limit)?

As for the rounding, keep the expression until the last step, and then plug in the values. As I said, it is okay for part 1 but not for part 2. We can get to that once you fix the integral.

5. Nov 27, 2008

### Melawrghk

Oh! I get it!
ln(6)/-0.3 = -5.973, and thus I should have added 5.973 to the LHS. Gaaah. I'm just so used to leaving out the variable=0 term. AAAH.

Anyways, okay, now it makes sense I guess. Thanks!

6. Nov 27, 2008

### naresh

Good.

Now for the rounding part: It is better to not use the value for s you obtain in the first part, directly in your equation for the second part. Instead substitute the expression for s, and plug in the value of t. This should give you a better result.

What is even better is to assume in the original problem that v = A - Bs where A and B are constants. Work out both parts of the problem using A and B; and finally substitute the values of A and B. This will give you different results, even with a solar-powered calculator. In general, when you subtract two large numbers that are almost the same, calculators tend to give you shaky results.