How Can Derivatives Determine Speed in Physics Problems?

[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!

 

[erobz]

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?

 

[emily-]

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.

 

[emily-]

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

 

[erobz]

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]

then integrated both sides

Show us the steps

 

[emily-]

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

 

[emily-]

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

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!!

 

[BvU]

Integral of dv/a = 2/3√v - 2/3√3

Nope ! What is the integral of ${2\over 3} {dv\over\sqrt v}$ ?

 

[kuruman]

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-]

Nope ! What is the integral of ${2\over 3} {dv\over\sqrt v}$ ?

It should be (4√v)/3 right?

 

[BvU]

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

 

[emily-]

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

Yess, thanks!!

 

[BvU]

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

( sorry for intruding, @erobz)

$\$

 

[erobz]

( sorry for intruding, @erobz)

$\$

Don't be!

 

[Orodruin]

The issue has been solved, but …

a = (1.5 v^(1/2)) m/s.

… this cannot be true! The units are all over the place!

 

[haruspex]

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?

 

[Orodruin]

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".

 

[haruspex]

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)?

 

[Orodruin]

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”.