Silly calculus problem I can't seem to do

[Steven G]

A particle moves on a straight line so that its velocity at any time t is given by v=12sqrt(s), where s is the distance from the origin.
If s=1 when t=0, then, when t=1, s equals ???

My observations:
1) WOLOG we can assume that the particle is moving along the x axis.
2) s must be a function of t, ie s = s(t)
3) at t=0, the particle is 1 unit from the origin (s=1)

Nothing else that I do seems to help.
Please supply me with a leading hint.

Thanks for your time.

 

[mario99]

[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s}\frac{1}{\sqrt{s}} \ ds[/imath]

 

[Dr.Peterson]

A particle moves on a straight line so that its velocity at any time t is given by v=12sqrt(s), where s is the distance from the origin.
If s=1 when t=0, then, when t=1, s equals ???

My observations:
1) WOLOG we can assume that the particle is moving along the x axis.
2) s must be a function of t, ie s = s(t)
3) at t=0, the particle is 1 unit from the origin (s=1)

Nothing else that I do seems to help.
Please supply me with a leading hint.

Thanks for your time.

Did you try writing a differential equation?

 

[Steven G]

Did you try writing a differential equation?

No, I didn't see any. That is a nice hint but unfortunately it didn't help me.

 

[Steven G]

[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s}\frac{1}{\sqrt{s}} \ ds[/imath]

I think that the 12 should be 1/6. It seems to me (and I can always be wrong) that you took the derivative of v = 12sqrt(s) and then integrated both sides. Up to a constant, you would end up with what you started with.

 

[mario99]

I think that the 12 should be 1/6. It seems to me (and I can always be wrong) that you took the derivative of v = 12sqrt(s) and then integrated both sides. Up to a constant, you would end up with what you started with.

Dear professor Steven,

I did this:

[imath]\displaystyle v = \frac{ds}{dt}[/imath]


[imath]\displaystyle v \ dt = ds[/imath]


[imath]\displaystyle \int_{0}^{1} v \ dt = \int_{1}^{s} ds[/imath]


[imath]\displaystyle \int_{0}^{1} 12\sqrt{s} \ dt = \int_{1}^{s} ds[/imath]


[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s} \frac{1}{\sqrt{s}} \ ds[/imath]

 

[Steven G]

Dear professor Steven,

I did this:

[imath]\displaystyle v = \frac{ds}{dt}[/imath]


[imath]\displaystyle v \ dt = ds[/imath]

So you are saying that v = ds/dt which is 6/sqrt(s). However, I though that v was given to be 12sqrt(s).
That is my problem with your work.
Thanks for trying! I appreciate your help.

 

[Dr.Peterson]

A particle moves on a straight line so that its velocity at any time t is given by v=12sqrt(s), where s is the distance from the origin.
If s=1 when t=0, then, when t=1, s equals ???

And I did this:

[imath]\displaystyle\frac{ds}{dt}=12\sqrt{s},\;\;s(0)=1[/imath]​

 

[Dr.Peterson]

So you are saying that v = ds/dt which is 6/sqrt(s).

No, that would be [imath]\displaystyle\frac{dv}{ds}[/imath].

 

[Steven G]

No, that would be [imath]\displaystyle\frac{dv}{ds}[/imath].

Yes, of course!!

 

[Steven G]

It is strange how I didn't get that the position of the particle was exactly s(t).
Dr Peterson straightened me out on that point. Thanks!!

 

[Dr.Peterson]

If only everything were x and y, there would never be any confusion.

 

[Steven G]

If only everything were x and y, there would never be any confusion.

Yeah, but at this point that shouldn't matter.

 

[MaxMath]

Dear professor Steven,

I did this:

[imath]\displaystyle v = \frac{ds}{dt}[/imath]


[imath]\displaystyle v \ dt = ds[/imath]


[imath]\displaystyle \int_{0}^{1} v \ dt = \int_{1}^{s} ds[/imath]


[imath]\displaystyle \int_{0}^{1} 12\sqrt{s} \ dt = \int_{1}^{s} ds[/imath]


[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s} \frac{1}{\sqrt{s}} \ ds[/imath]

The last step is wrong. You cannot take [imath]\sqrt{s}[/imath] out of the integral because it's a function of [imath]s[/imath].

 

[MaxMath]

[math]s=1+\int_0^1{v}dt[/math][math]v=12\sqrt{s}[/math][math]v=12\sqrt{1+\int_0^1{v}dt}[/math]Solve [imath]v=v(t)[/imath] from the last line (also using the initial condition [imath]v=12[/imath] when [imath]t=0[/imath]), then use the first line to get [imath]s[/imath] at [imath]t=1[/imath].

 

[khansaheb]

And I did this:

\(\displaystyle \displaystyle\frac{ds}{dt}=12\sqrt{s},\;\;s(0)=1\)​

Then

\(\displaystyle \displaystyle\frac{ds}{\sqrt{s}}=12{dt},\;\;\)

\(\displaystyle {s}^{\frac{1}{2}} + C =12{t},\;\;\) where C = -1

\(\displaystyle {s}^{\frac{1}{2}} - 1 =12{t},\;\;\)

When t = 1

s = 169 (unit is not given)

 

[MaxMath]

[math]s=1+\int_0^1{v}dt[/math][math]v=12\sqrt{s}[/math][math]v=12\sqrt{1+\int_0^1{v}dt}[/math]Solve [imath]v=v(t)[/imath] from the last line (also using the initial condition [imath]v=12[/imath] when [imath]t=0[/imath]), then use the first line to get [imath]s[/imath] at [imath]t=1[/imath].

I made some mistakes. It should read:

[math]s=1+\int_0^t{v}dt[/math][math]v=12\sqrt{s}[/math][math]v=12\sqrt{1+\int_0^t{v}dt}[/math]Solve [imath]v=v(t)[/imath] from the last line (also using the initial conditions [imath]s=1[/imath] and [imath]v=12[/imath] when [imath]t=0[/imath]), then use the first line to get [imath]s[/imath] at [imath]t=1[/imath].

 

[mario99]

The last step is wrong. You cannot take [imath]\sqrt{s}[/imath] out of the integral because it's a function of [imath]s[/imath].

I am bad at math. Thanks for noticing that!

 

[Dr.Peterson]

I am bad at math. Thanks for noticing that!

Actually, you weren't entirely wrong; you just wrote the integral too soon. I'll correct your work:

I did this:

[imath]\displaystyle v = \frac{ds}{dt}[/imath]

[imath]\displaystyle v \ dt = ds[/imath]

[imath]\displaystyle 12\sqrt{s} \ dt = ds[/imath]

[imath]\displaystyle 12 \ dt = \frac{1}{\sqrt{s}} \ ds[/imath]

[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s} \frac{1}{\sqrt{s}} \ ds[/imath]

It's okay to move the radical when it isn't yet inside an integral.

Now finish:

[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s} \frac{1}{\sqrt{s}} \ ds[/imath]​

​

[imath]\displaystyle \left.12t\right|_{0}^{1}=\left.2s^{1/2}\right|_{1}^{s}[/imath]​

​

[imath]\displaystyle 12=2s^{1/2}-2[/imath]​

​

[imath]\displaystyle 6=s^{1/2}-1[/imath]​

​

[imath]\displaystyle s=(6+1)^2=49[/imath]​

I hadn't noticed @khansaheb's small error.

 

[MaxMath]

\(\displaystyle \displaystyle\frac{ds}{\sqrt{s}}=12{dt},\;\;\)

\(\displaystyle {s}^{\frac{1}{2}} + C =12{t},\;\;\) where \(\displaystyle C = -1[\tex]

\(\displaystyle {s}^{\frac{1}{2}} - 1 =12{t},\;\;\)

When t = 1

s = 169 (unit is not given)
\)

I think you missed a 2 on the left side?

\(\displaystyle \displaystyle\frac{ds}{\sqrt{s}}=12{dt},\;\;\)

\(\displaystyle 2{s}^{\frac{1}{2}} + C =12{t},\;\;\) where C = -2

\(\displaystyle 2{s}^{\frac{1}{2}} - 2=12{t}\;\;\)

\(\displaystyle s=(\frac{12{t}+2}{2})^2\;\;\)

When t = 1

s = 49