# Homework Help: Drag force deceleration

1. Sep 30, 2016

### nysnacc

1. The problem statement, all variables and given/known data

I forgot if the deceleration is -0.003v or same here -0.003v2

2. Relevant equations
v = u + at (first acceleration)
dv/dt = a (deceleratioin) = -0.003v^2

3. The attempt at a solution
From first equation,
342.2 = 0 + 30*t1
t1= 11.407s

From 2nd equation,
dv/v2 = -0.003
[-1/3.422] - [-1/342.2] = -0.003 (t2-t1)

------------------------------------------

If the decelerate is -0.003v
dv/v = -0.003
ln (3.422) - ln (342.2) = -0.003 (t2-t1)

t2 (total time) = 1544.46 s

Is the answer making sense?

2. Sep 30, 2016

### nysnacc

Updates: should be -0.003 v

3. Oct 1, 2016

### ehild

Correct, so what is t2, the total time?
------------------------------------------
In the problem text, the deceleration is a = -0.003 v2, and you did not calculate the time. In case of a= -0.003v, your answer is correct. Use four significant digits.

4. Oct 2, 2016

### nysnacc

thanks for reply. The question was using a = -0.003v as deceleration.
And my t is 1544.46s, which is 25 min, is it making sense? cuz it is soooo long.

5. Oct 2, 2016

### ehild

Yes, it is correct, with that low deceleration, which is linear in velocity. At that high speed, quadratic deceleration would be more realistic

6. Oct 2, 2016

### nysnacc

Okay thanks, I double checked the question, it uses v instead of v2, which you mentioned to be more realistic.

7. Oct 2, 2016

### haruspex

On a pedantic note, the question should not give the acceleration as either -0.003v m/s2 or -0.003v2 m/s2.
Rather, the units should be given as s-1 and m-1 respectively. This is because v will come with its own units. The units specified in the expression for a should only be the units of the -0.003 constant.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted