How Can Derivatives Determine Speed in Physics Problems?

  • #1
emily-
15
2
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!
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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
 
  • Like
Likes erobz
  • #4
Yes, sub in the differential form for ##a##, separate variables and integrate each side.
 
  • #5
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
 
  • #6
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
 
  • Like
Likes emily-
  • #7
emily- said:
then integrated both sides
Show us the steps
 
  • #8
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
 
  • #9
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!!
 
  • Like
Likes erobz
  • #10
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} ## ?
 
  • #11
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.
 
  • Like
Likes emily-
  • #12
BvU said:
Nope ! What is the integral of ##{2\over 3} {dv\over\sqrt v} ## ?
It should be (4√v)/3 right?
 
  • #13
Yes. So you get ##\sqrt {v(t)}-\sqrt{v(0)} = {3\over 4} t##. Move to the other side and square.
 
  • Like
Likes emily-
  • #14
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!!
 
  • #15
You are welcome. And ##\LaTeX## really is fun !

(:smile: sorry for intruding, @erobz)

##\ ##
 
  • Like
Likes emily-
  • #16
BvU said:
(:smile: sorry for intruding, @erobz)

##\ ##
Don't be!
 
  • Like
Likes BvU
  • #17
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!
 
  • Like
Likes SammyS and erobz
  • #18
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?
 
  • #19
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".
 
  • #20
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)?
 
  • #21
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”.
 

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
243
  • Introductory Physics Homework Help
Replies
6
Views
973
  • Introductory Physics Homework Help
Replies
17
Views
622
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
849
  • Introductory Physics Homework Help
Replies
8
Views
309
  • Introductory Physics Homework Help
Replies
7
Views
3K
Back
Top