# How Can Derivatives Determine Speed in Physics Problems?

• emily-
emily-
Homework Statement
The acceleration of the boat is defined by
a = (1.5 v^(1/2)) m/s. Determine its speed when t = 4 s if it has
a speed of 3 m/s when t = 0.
Relevant Equations
v = ds/dt
a = dv/dt
a ds = v dv
I have been trying to solve this problem for hours using the mentioned equations but no matter what I do I cannot get the correct answer, that is v = 22.4 m/s. I thought that maybe if I could get an expression where v is a function of time I could solve the problem but I don't know how to do that since they always end up being separate. Any help and guidance would be deeply appreciated, thanks!

One of your "relevant equations" is capable of solving your issue. You are given:

$$a = 1.5 \sqrt{v}$$

Which one do you think it is?

erobz said:
One of your "relevant equations" is capable of solving your issue. You are given:

$$a = 1.5 \sqrt{v}$$

Which one do you think it is?
a = dv/dt

erobz
Yes, sub in the differential form for ##a##, separate variables and integrate each side.

erobz said:
Yes, sub in the differential form for ##a##, separate variables and integrate each side.
I rewrote the equation to: dt = dv/a and then integrated both sides. When it comes to the boundaries, I wrote 4 <- 0 for dt and v <- 3 for dv/a but as a result I got 0.023 m/s

You're having some computational error. I get ##22.4 ~\rm{m/s}## when I do it. Please show your work using Latex. See: LaTeX Guide

emily-
emily- said:
then integrated both sides
Show us the steps

Integral of dt = 4
Integral of dv/a = 2/3√v - 2/3√3
I move the numerical expressions to one side and isolate √v and square it

erobz said:
You're having some computational error. I get ##22.4 ~\rm{m/s}## when I do it. Please show your work using Latex. See: LaTeX Guide
BvU said:
Show us the steps
I got the integral expression wrong!! It's (2√v)/1.5 and not 2/3√v ! Omg
Thanks for the help!!

erobz
emily- said:
Integral of dv/a = 2/3√v - 2/3√3
Nope ! What is the integral of ##{2\over 3} {dv\over\sqrt v} ## ?

emily- said:
I rewrote the equation to: dt = dv/a and then integrated both sides. When it comes to the boundaries, I wrote 4 <- 0 for dt and v <- 3 for dv/a but as a result I got 0.023 m/s
Please show you work. I suspect you made a mistake when you integrated. Consider using LaTeX to post your equations. It makes them more legible and it's a good skill to have. Click on the link "LaTeX Guide", lower left to see how it is done.

emily-
BvU said:
Nope ! What is the integral of ##{2\over 3} {dv\over\sqrt v} ## ?
It should be (4√v)/3 right?

Yes. So you get ##\sqrt {v(t)}-\sqrt{v(0)} = {3\over 4} t##. Move to the other side and square.

emily-
BvU said:
Yes. So you get ##\sqrt {v(t)}-\sqrt{v(0)} = {3\over 4} t##. Move to the other side and square.
Yess, thanks!!

You are welcome. And ##\LaTeX## really is fun !

( sorry for intruding, @erobz)

##\ ##

emily-
BvU said:
( sorry for intruding, @erobz)

##\ ##
Don't be!

BvU
The issue has been solved, but …
emily- said:
a = (1.5 v^(1/2)) m/s.
… this cannot be true! The units are all over the place!

SammyS and erobz
Orodruin said:
The issue has been solved, but …

… this cannot be true! The units are all over the place!
Right. Should be ##(1.5 v^{\frac 12}) m^{\frac 12}s^{-\frac 32}##, yes?

haruspex said:
Right. Should be ##(1.5 v^{\frac 12}) m^{\frac 12}s^{-\frac 32}##, yes?
Yes, if ##v## is dimensionful with the appropriate units. However, sometimes you see introductory texts introduce dimensionless quantities such as "velocity is ##v## m/s", making ##v## actually dimensionless. This looks like some strange combination, but that's how I would do it (although I would probably write it as ##(1.5~{\rm m}^{1/2}/{\rm s}^{3/2}) \sqrt{v}## or - even better - ##k\sqrt v##, where ##k = 1.5~{\rm m}^{1/2}/{\rm s}^{3/2}##).

If one wants to use dimensionless quantities here, I would have formulated it just like "##a = 1.5\sqrt{v}##, where the acceleration is ##a## m/s2 and the velocity is ##v## m/s".

Orodruin said:
Yes, if v is dimensionful with the appropriate units.
Aren't the units of v irrelevant to the correct form (as long as it has the dimension of velocity)?

haruspex said:
Aren't the units of v irrelevant to the correct form (as long as it has the dimension of velocity)?
Sure. With ”suitable” I only mean ”of the correct dimensions”.

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