# Calculus involving accln and vel

## 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

AlephZero
Science Advisor
Homework Helper
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
Science Advisor
Homework Helper
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
$$v^{-\frac{1}{2}}dv= -\alpha dt[/itex] That should be easy to integrate. Once you have solved for v(t), of course [tex]v= \frac{dx}{dt}$$
so you have a second integration for x(t).

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
$$v^{-\frac{1}{2}}dv= -\alpha dt[/itex] That should be easy to integrate. Once you have solved for v(t), of course [tex]v= \frac{dx}{dt}$$
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:

$$2 \sqrt{v} = \alpha t$$

<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?????

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
Science Advisor
Homework Helper
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, $t = \frac {2 \sqrt{v(0)}}{\alpha}$ 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 $\frac{dx}{dt}= v= (-\sqrt{v(0)}-\frac{\alpha}{2}t)^2$.

is this true??

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

is that right???

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 isnt a mess..... 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
Science Advisor
Homework Helper
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,
[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).