Asked by Stephen
                The Question: A particle moves along the X-axis so that at time t > or equal to 0 its position is given by x(t) = cos(√t). What is the velocity of the particle at the first instance the particle is at the origin?
So far I was able to determine that the velocity of the particle would be undefined but, I don't understand what it means in the context of this problem for velocity to be undefined. What is the particle doing if its velocity is undefined?
            
        So far I was able to determine that the velocity of the particle would be undefined but, I don't understand what it means in the context of this problem for velocity to be undefined. What is the particle doing if its velocity is undefined?
Answers
                    Answered by
            bobpursley
            
    x=cos(sqrt(t))
v=-sin(sqrt(t))*1/2sqrtt
wo when is position zero?
0=cos(sqrt(t))
sqrt(t)=PI/2
t=PI^2/4
v(PI^2/4)=-sin(PI/2)*1/2sqrtPI/2)
but sin(PI/2)=1
v(PI^2/4)=1/sqrt(PI/2)= 0.797884561
So I dont see where the undefined comes from.
    
v=-sin(sqrt(t))*1/2sqrtt
wo when is position zero?
0=cos(sqrt(t))
sqrt(t)=PI/2
t=PI^2/4
v(PI^2/4)=-sin(PI/2)*1/2sqrtPI/2)
but sin(PI/2)=1
v(PI^2/4)=1/sqrt(PI/2)= 0.797884561
So I dont see where the undefined comes from.
                    Answered by
            Stephen
            
    I believe your equation for velocity is incorrect.
In this case
V=-sin(sqrt(t))*1/2(sqrt(t))^-1/2
Your equation for velocity is missing raising to the -1/2 power at the end. applying that piece will make velocity undefined at t=0.
    
In this case
V=-sin(sqrt(t))*1/2(sqrt(t))^-1/2
Your equation for velocity is missing raising to the -1/2 power at the end. applying that piece will make velocity undefined at t=0.
                    Answered by
            bobpursley
            
    Nope. 
x= cos(t^1/2)
v=-sin(t^1/2) *1/2*1/t^1/2
= -sin(sqrtt)*1/(2sqrtt) which is what I have. Now, how does having
sqrt(PI/2) in the denominator make it undefined. I must not be seeing your point.
 
    
x= cos(t^1/2)
v=-sin(t^1/2) *1/2*1/t^1/2
= -sin(sqrtt)*1/(2sqrtt) which is what I have. Now, how does having
sqrt(PI/2) in the denominator make it undefined. I must not be seeing your point.
                    Answered by
            Stephen
            
    Oh lol, I think I see what we're doing differently.
I thought that the first instance that the particle would be at the origin was when t=0 but it really is when t=pi/2.
Basically, I substituted the wrong number.
Thank you, and I'm sorry for not seeing your point.
    
I thought that the first instance that the particle would be at the origin was when t=0 but it really is when t=pi/2.
Basically, I substituted the wrong number.
Thank you, and I'm sorry for not seeing your point.
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