How to find kinetic energy of a particle as a function of time?

[wcwcaseman]

Homework Statement

The position of a 5kg particle moving in the x-y plane varies as a function of time according to the expression

r = [(3t + 1)m]i + [(-4t^2 +5t + 2)m]j

(r is a vector I just didn't know how to make the arrow over head)


Find the kinetic energy of the particle as a function of time

Homework Equations

k = 1/2 mx+b

The Attempt at a Solution

d/dx r = [(3t + 1)m]i + [(-4t^2 +5t + 2)m]j

v = 3i + (-8t +5)j

v =(3i^2 + (-8t +5)j^2)^(1/2)?

.........

3t+1 = 0
t= 1/3 for x

-4t^2 +5t + 2

t= -.44 or t= 1.69 for y??

Very confused don't know what vector r is. Is it already a function of time or what. I'm very confused and don't understand how to solve please help

 

[rock.freak667]

KE=1/2 mv²

and v² is the same as v.v

 

[wcwcaseman]

KE=1/2 mv²

and v² is the same as v.v

so ...

v = 3i * (-8t +5)j

v = (-24t +15)^1/2? and if so were do I go from there?

 

[rock.freak667]

so ...

v = 3i * (-8t +5)j

v = (-24t +15)^1/2? and if so were do I go from there?

If v = 3i + (-8t +5)j, then compute v.v, the dot product of v and v.

 

[wcwcaseman]

If v = 3i + (-8t +5)j, then compute v.v, the dot product of v and v.

I'm not sure if I'm correct in my assumption that v = 3i + (-8t +5)j also I don't see how the dot product could be used since we have the variable t ?

 

[ideasrule]

What was that "m" in your initial equation for r? If that shouldn't have been there, your expression for v is correct.

Who cares about the t. Just do v dot v the usual way and see what happens. Do you get something with t in it? That's how v^2 varies with t.