Dynamics: Curvilinear motion

[chrsr34]

Hi guys,Im Chris, I am new here. I am in Dynamics doing curvilinear motion.

Homework Statement

A truck travels at a speed of 4m/s along a circular road with radius if 50m. For a short distance, from s=0, its speed is then increased by a(tan) = (0.05s) m/s, where s is in meters. Determine the speed and acceleration magnitude at s=10. (i can find the a-magnitude easily after i find v, i just can't find v)

Homework Equations

acceleration and velocity equations
a(tan)ds = v dv

The Attempt at a Solution

Ive tried to attempt this using integration with respect to s, but this does not yield the correct answer, so its the incorrect procedure. The answer is 4.58 m/s.
I have never seen or dealt with integration with respect to displacement, therefore i don't know how to approach this problem. The book says nothing on figuring this out. Any help is appreciated, Thanks!
Chris

 

[quantumdude]

Hello, and welcome to PF!

Integration is indeed the way to go, so you must have made an error there. Let's see what you did, and then we can help you.

 

[chrsr34]

Thank you! well i integrated and got v = 0.025s^2, s = 10. This yeilds 2.5 m/s. Adding this to the initial 4 m/s yields 6.5 m/s. This is incorrect. the answer is 4.58 m/s. So, where is my error? I am quite dumbfounded...

 

[chrsr34]

Im starting to think this should be in advanced physics as it requires perhaps some difficult manipulation? I am in calc 3 and have no idea what to do...
To the Moderator, please move if you feel fit

 

[Rainbow Child]

Thank you! well i integrated and got v = 0.025s^2, s = 10. This yeilds 2.5 m/s. Adding this to the initial 4 m/s yields 6.5 m/s. This is incorrect. the answer is 4.58 m/s. So, where is my error? I am quite dumbfounded...

You can not add the initial velocity. From the integration you have

$$\int_4^{v_f} v\,d\,v=\int_0^{s_f}\frac{s}{20}d\,s\Rightarrow \frac{1}{2}\,v^2\Big|_4^{v_f}=\frac{1}{40}\,s^2_f \Rightarrow u_f^2-4^2=\frac{1}{20}\,s_f^2$$

Plug your values in the above equation to find the answer.

 

[chrsr34]

wow,i did not even think about integrating the v from 4-vf. Thank you so much. But just to be clear, you say that i can't add the initial velocity. The final answer of that integration is .58m/s so that would conclude that I do have to add the initial 4 m/s to get 4.58m/s. Isnt this correct?

 

[Rainbow Child]

The final answer of that integration is .58m/s

Which integration?
If the initial velocity is $v_i$ then the final velocity is $v_$ then

$$v_f^2=v_i^2+\frac{1}{20}\,s^2\Rightarrow v_f=\sqrt{v_i^2+\frac{1}{20}\,s^2}$$

The final velocity is not proportional to the initial velocity. That's why you can not add it.

Is it clear?

 

[chrsr34]

Hmm, now I am even more confused. Plugging s and vi into that equation gives me 6.4 m/s = vf. I don't understand, if the initial vi is 4m/s, and then accelerates for 10m to a new vf, wouldn't you be adding vi + delta v to get final velocity?

$$u_f^2-4^2=\frac{1}{20}\,s_f^2$$ solving for uf yields .58
adding 4+.58 yeilds the correct answer

 

[Rainbow Child]

Hmm, now I am even more confused. Plugging s and vi into that equation gives me 6.4 m/s = vf. I don't understand, if the initial vi is 4m/s, and then accelerates for 10m to a new vf, wouldn't you be adding vi + delta v to get final velocity?

How did you find $v_f=6.4\,m/sec$? For $s=10\,m,\,v_i=4\,m/sec$ you get

$$v_f=\sqrt{16+\frac{100}{20}}\Rightarrow v_f=\sqrt{21}\Rightarrow v_f=4.58\,m/sec$$

The definition of the acceleration is $a=\frac{d\,v}{d\,t}\Rightarrow d\,v=a\,d\,t\Rightarrow v_f-v_i=\int_0^{t_f}a\,d\,t$. If you call $\Delta v=\int_0^{t_f}a\,d\,t$ then of course you can write $v_f=v_i+\Delta v$.

 

[chrsr34]

oh yes, your right, i was squaring the 100/20 by accident. I still don't fully understand the entire process but ill try and figure it out. Thanks again Rainbow Child.