Calculus involving accln and vel

[sidrox]

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 traveled before it stops
iii. The time at which the instantaneous velocity is (v1)/4
iv. Distance traveled 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...

 

[AlephZero]

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.

 

[HallsofIvy]

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
\frac{dv}{dt}= -\alpha \sqrt{v}= \alpha v^{\frac{1}{2}}

That's a "separable" equation- you can write it as
v^{-\frac{1}{2}}dv= -\alpha dt[/itex]<br /> <br /> That should be easy to integrate. Once you have solved for v(t), of course<br /> v= \frac{dx}{dt}<br /> so you have a second integration for x(t).

 

[sidrox]

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
\frac{dv}{dt}= -\alpha \sqrt{v}= \alpha v^{\frac{1}{2}}

That's a "separable" equation- you can write it as
v^{-\frac{1}{2}}dv= -\alpha dt[/itex]<br /> <br /> That should be easy to integrate. Once you have solved for v(t), of course<br /> v= \frac{dx}{dt}<br /> so you have a second integration for x(t).

<br /> <br /> Hmmm...i got the idea...<br /> but my problem now is how to apply the limits...on integrating, what i got was:<br /> <br /> 2 \sqrt{v} = \alpha t <br /> <br /> &lt;i hope the latex is right, i am using it for the first time!<br /> <br /> now for the above equation, what will my limits be and where do i bring in the v1 term?

 

[sidrox]

Progress!

- \alpha \sqrt{v(t)}= \frac {dv}{dt}

- \alpha t = 2 \sqrt{v(t)} + 2\sqrt{v(0)}

- \alpha t = \{2 \sqrt{v(0)}

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

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

Now how do i proceed?

 

[HallsofIvy]

i. Time taken by the particle to come to rest
ii. The distance traveled before it stops
iii. The time at which the instantaneous velocity is (v1)/4
iv. Distance traveled at the time the velocity becomes (v1)/4.

Yes, t = \frac {2 \sqrt{v(0)}}{\alpha} is the time until the particle comes to rest (v= 0).
Now, since you know that - \alpha t = 2 \sqrt{v(t)} + 2\sqrt{v(0)} you can solve that equation for v(t): 2\sqrt{v(t)}= -2\sqrt{v(0)}- \alpha t so \frac{dx}{dt}= v= (-\sqrt{v(0)}-\frac{\alpha}{2}t)^2.

 

[sidrox]

is this true??

\frac{d}{dt} \-v(0) = 0?

is that right?

 

[sidrox]

okay...here is my working:

2 \sqrt{v} = - 2\sqrt{v(0)} - \frac {\alpha t}{2}

On squaring,

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

On writing v as \frac {dx}{dt}, and then integrating:

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

and i am going to continue in a separate post becos i am new to laTex and i hope all that i have typed isn't a mess...

 

[sidrox]

Now the body comes to rest at

t = \frac {2 \sqrt{v(0)}}{\alpha}

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:

x = \frac {8 \sqrt{v(0)}^3}{3 \alpha}


Where am I going wrong?

 

[HallsofIvy]

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
2\sqrt{v(t)}= -\alpha t+ C
Taking t= 0 2\sqrt{v1}= C. That gives
2\sqrt{v(t)}= 2\sqrt{v1}- \alpha t
squaring that equation,
4 v(t)= 4 v1- 4\sqrt{v1}\alpha t+ \alpha^2 t^2[/itex]<br /> <br /> The particle will &quot;come to rest&quot; when that is equal to 0.<br /> <br /> Now you have<br /> v(t)= \frac{dx}{dt}= v1- \sqrt{v1}\alpha t+ \frac{\alpha^2}{4} t^2[/itex]&lt;br /&gt; &lt;br /&gt; That should be easy to integrate for x(t).