Find Velocity for Parametric Equations with t = 2s

[the pro]

Homework Statement

the parametric equations of the motion are x=(t)^2m and y=(6t-1.5t^2)m i need to find the
velocity at the time moment of t=2s

Homework Equations

The Attempt at a Solution

so this are the four solutions v=o,v=2m/s,v=4m/s,6m/s .

 

[korican04]

Homework Statement

the parametric equations of the motion are x=(t)^2m and y=(6t-1.5t^2)m i need to find the
velocity at the time moment of t=2s

Homework Equations

The Attempt at a Solution

so this are the four solutions v=o,v=2m/s,v=4m/s,6m/s .

Do you know how to take a derivative?
dx/dt = 2t m/s
dy/dt= 6-3t m/s

At t=2
dx/dt=4 m/s
dy/dt=0 m/s

v= 4 m/s in the x direction.

 

[the pro]

No i don't now how if you could post it how to do also the derivate then it would be ok and thanks for this post

 

[Astronuc]

No i don't now how if you could post it how to do also the derivate then it would be ok and thanks for this post

See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/vel2.html

One has two position parameters x(t) and y(t).

The velocity, or rather speed, in each direction is just the first derivative with respect to time,

v_x(t) = dx(t)/dt, and v_y(t) = dy(t)/dt.

http://hyperphysics.phy-astr.gsu.edu/hbase/deriv.html
http://hyperphysics.phy-astr.gsu.edu/hbase/deriv.html#c3

Then since velocity is a vectors,

v(t) = v_x(t) i + v_y(t) j, where i and j are just the unit vectors in x and y directions.

The magnitude of v(t) is given by the square root of the sum of the squares of the speeds in both direction, i.e.,

|v(t)| = sqrt (v_x²(t) + v_y²(t))

See also - http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html

http://hyperphysics.phy-astr.gsu.edu/hbase/vsca.html#vsc1

 

[the pro]

Thanks very much sir astronuc