Vf=Vi+at
Vf=9j+(2i-4j)t
displacement= origposition+at
now, in the i direction
15=O+2t or t=7.5 when x position is 15
put that into the first equation, solve for Vf.
speed= sqrt (xcomponent^2+ycomponent^2)
you get the x component, y component in the Vf equation
3. At t = 0 , a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with constant accelaration of ( 2.0i - 4.0j ) m/s2 . At the instant the x coordinate of the particle is 15 m , what is the speed of the particle ?
1 answer