Deceleration Calculation for Drag Force with Varying Units

[nysnacc]

Homework Statement

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

Homework Equations

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

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)

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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?

 

[nysnacc]

Updates: should be -0.003 v

 

[ehild]

Homework Statement

View attachment 106758

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

Homework Equations

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

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)

Correct, so what is t2, the total time?
------------------------------------------

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?

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.

 

[nysnacc]

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.

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? because it is soooo long.

 

[ehild]

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? because it is soooo long.

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

 

[nysnacc]

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

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

 

[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.