Question

the position of a particle moving in x-y plane is given as
x=t4-t3+6t2+5
y=t3-3t2+4t+9
a)find velocity ,its magnitude&direction just after t=2s.

Answer 1

differentiate both equations to get the general equation velocities in each axis. put in t=2 to find out the 'x' and 'y' components of the velocity.

magnitude will be sqrt of sum square of velocity component in each axis.
so if its Vx and Vy, then
\[|V| = \sqrt{V_x^2 + V_y^2}\]

direction of velocity is, suppose angle 'A' from the positive X-axis.

\[so,A=tan^{-1}(\frac{V_y}{V_x})\]

Answer 2

yes boss well said

Answer 3

thnx @apoorvk