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Calculus involving accln and vel

  • Thread starter sidrox
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  • #1
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Homework Statement


A point moves rectilinearly with a deceleration whose modulus depends on the velocity v of the particle as

a=(alpha) X (square root of velocity v),​

where alpha is a positive constant. At time t=0, the particle has initial velocity v1.

Then, what are the following:
i. Time taken by the particle to come to rest
ii. The distance travelled before it stops
iii. The time at which the instantaneous velocity is (v1)/4
iv. Distance travelled at the time the velocity becomes (v1)/4.

The Attempt at a Solution


I tried to integrate the equation but ended up no where because of the v term in the given equation, which kept getting in my way...PLEASE HELP!!
I want to learn how to solve such problems which involve such variables in their equations.....
 

Answers and Replies

  • #2
AlephZero
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Writing x' and x" for dx/dt and d^2/dt^2 (Latex takes too long!)

Your equation is x" = ax'^2

The "trick" for solving this is to substitute y = (1/2)x'^2

y' = x'x"
x" = y'/x' = (dy/dt) / (dx/dt) = dy/dx

So the equation becomes dy/dx = 2ay.
 
  • #3
HallsofIvy
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No, the equation is NOT x"= ax'^2: it is x"= -a sqrt(x').
(negative a (or alpha) because you are told that the deceleration is given by alpha sqrt(v) where alpha is a positive number.)

I would have done it just slightly differently. Your base equation is
[tex]\frac{dv}{dt}= -\alpha \sqrt{v}= \alpha v^{\frac{1}{2}}[/tex]

That's a "separable" equation- you can write it as
[tex]v^{-\frac{1}{2}}dv= -\alpha dt[/itex]

That should be easy to integrate. Once you have solved for v(t), of course
[tex]v= \frac{dx}{dt}[/tex]
so you have a second integration for x(t).
 
  • #4
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No, the equation is NOT x"= ax'^2: it is x"= -a sqrt(x').
(negative a (or alpha) because you are told that the deceleration is given by alpha sqrt(v) where alpha is a positive number.)

I would have done it just slightly differently. Your base equation is
[tex]\frac{dv}{dt}= -\alpha \sqrt{v}= \alpha v^{\frac{1}{2}}[/tex]

That's a "separable" equation- you can write it as
[tex]v^{-\frac{1}{2}}dv= -\alpha dt[/itex]

That should be easy to integrate. Once you have solved for v(t), of course
[tex]v= \frac{dx}{dt}[/tex]
so you have a second integration for x(t).
Hmmm....i got the idea....
but my problem now is how to apply the limits...on integrating, what i got was:

[tex] 2 \sqrt{v} = \alpha t[/tex]

<i hope the latex is right, i am using it for the first time!!

now for the above equation, what will my limits be and where do i bring in the v1 term?????
 
  • #5
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Progress!!!

[tex] - \alpha \sqrt{v(t)}= \frac {dv}{dt} [/tex]

[tex] - \alpha t = 2 \sqrt{v(t)} + 2\sqrt{v(0)} [/tex]

[tex] - \alpha t = \{2 \sqrt{v(0)} [/tex]

The above is because the velocity is 0 (in the first sub division of the question)

and thus,
[tex] t = \frac {2 \sqrt{v(0)}}{\alpha} [/tex]

Now how do i proceed???
 
  • #6
HallsofIvy
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i. Time taken by the particle to come to rest
ii. The distance travelled before it stops
iii. The time at which the instantaneous velocity is (v1)/4
iv. Distance travelled at the time the velocity becomes (v1)/4.
Yes, [itex] t = \frac {2 \sqrt{v(0)}}{\alpha} [/itex] is the time until the particle comes to rest (v= 0).
Now, since you know that [itex] - \alpha t = 2 \sqrt{v(t)} + 2\sqrt{v(0)} [/itex] you can solve that equation for v(t): [itex]2\sqrt{v(t)}= -2\sqrt{v(0)}- \alpha t[/itex] so [itex]\frac{dx}{dt}= v= (-\sqrt{v(0)}-\frac{\alpha}{2}t)^2[/itex].
 
  • #7
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is this true??

[tex]\frac{d}{dt} \-v(0) = 0 [/tex]????

is that right???
 
  • #8
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okay...here is my working:

[tex] 2 \sqrt{v} = - 2\sqrt{v(0)} - \frac {\alpha t}{2} [/tex]

On squaring,

[tex] v = v(0) + \frac {\alpha^2 t^2}{4} + \sqrt{v(0)}\alpha t [/tex]

On writing v as [tex] \frac {dx}{dt} [/tex], and then integrating:

[tex]x = 0 + \frac {\alpha^2 t^3}{12} + \frac { \sqrt{v(0)} \alpha t^2}{2} [/tex]

and i am going to continue in a separate post becos i am new to laTex and i hope all that i have typed isnt a mess.....:rolleyes:
 
  • #9
22
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Now the body comes to rest at

[tex] t = \frac {2 \sqrt{v(0)}}{\alpha} [/tex]

Thus, i substituted the above expression for 't' in the expression for 'x'.

and i have gotten the following value for x, but it is not one of the options.

i got:

[tex] x = \frac {8 \sqrt{v(0)}^3}{3 \alpha} [/tex]


Where am I going wrong?????????
 
  • #10
HallsofIvy
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You were NOT told that the initial velocity was 0! You were told
At time t=0, the particle has initial velocity v1.
You have determined that
[tex]2\sqrt{v(t)}= -\alpha t+ C[/tex]
Taking t= 0 [itex]2\sqrt{v1}= C[/itex]. That gives
[tex]2\sqrt{v(t)}= 2\sqrt{v1}- \alpha t[/tex]
squaring that equation,
[tex]4 v(t)= 4 v1- 4\sqrt{v1}\alpha t+ \alpha^2 t^2[/itex]

The particle will "come to rest" when that is equal to 0.

Now you have
[tex]v(t)= \frac{dx}{dt}= v1- \sqrt{v1}\alpha t+ \frac{\alpha^2}{4} t^2[/itex]

That should be easy to integrate for x(t).
 

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