# Calculus involving accln and vel

• sidrox
You can also find the distance traveled when the particle stops by integrating $$v(t)= v1- \sqrt{v1}\alpha t+ \frac{\alpha^2}{4} t^2[/itex]from 0 to t when v(t)= 0.I want the time when v= v1/4 and the distance traveled at that time.I'm sorry, you did specify v1/4. In that case, you need to solvev1/4= v1- \sqrt{v1}\alpha t+ \frac{\alpha^2}{4} t^2That is a quadratic equation in t. You can solve that using the quadratic #### sidrox ## Homework Statement A point moves rectilinearly with a deceleration whose modulus depends on the velocity v of the particle as a=(alpha) X (square root of velocity v),​ where alpha is a positive constant. At time t=0, the particle has initial velocity v1. Then, what are the following: i. Time taken by the particle to come to rest ii. The distance traveled before it stops iii. The time at which the instantaneous velocity is (v1)/4 iv. Distance traveled at the time the velocity becomes (v1)/4. ## The Attempt at a Solution I tried to integrate the equation but ended up no where because of the v term in the given equation, which kept getting in my way...PLEASE HELP! I want to learn how to solve such problems which involve such variables in their equations... Writing x' and x" for dx/dt and d^2/dt^2 (Latex takes too long!) Your equation is x" = ax'^2 The "trick" for solving this is to substitute y = (1/2)x'^2 y' = x'x" x" = y'/x' = (dy/dt) / (dx/dt) = dy/dx So the equation becomes dy/dx = 2ay. No, the equation is NOT x"= ax'^2: it is x"= -a sqrt(x'). (negative a (or alpha) because you are told that the deceleration is given by alpha sqrt(v) where alpha is a positive number.) I would have done it just slightly differently. Your base equation is [tex]\frac{dv}{dt}= -\alpha \sqrt{v}= \alpha v^{\frac{1}{2}}$$

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).

HallsofIvy said:
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?

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 $\frac{dx}{dt}= v= (-\sqrt{v(0)}-\frac{\alpha}{2}t)^2$.

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 isn't 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?

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).