A particle movesin the X-Y plane with a constasnt acceleration of `1.5 m/s^2` in the direction making an angle of `37^0` with the X-axis.At t=0 the particle is at the orign and its velocity is 8.0 m/s along the X-axis. Find the velocity and the position of the particle at t=4.0 s.

Explanation

`a_x=(1.5 m/s^2) (cos37^0)`
`=(1.5 m/s^2)xx4/5 = 1.2 m/s^2`
and a_y=(1.5 m/s^2)(sin37^0)`
`=(1.5 m/s^2) xx 3/5 = 0.90 m/s^2`. ltbr. The initial velocity has components
`u_x=8.0 m/s`
and `u_y=0`
`At t=0, x=0 and y=-0`
The x-component of the velocity at time t=4.0 s is given by
`v_x=u_x+a_xt`
`=8.0 m/s+(1.2 m/s^2)(4.0s)`
`=8.0 m/s+4.8 m/s=12.8 m/s`.
The y-componet of velocity at t=4.0 si given by
`v_y=u_y+a_y t`
`=0+(0.90 m/s^2)(4.0s)=3.6 m/s`
The velocity of the particle ast t=4.0 is
`v=sqrt(v_x^2+v_y^2)= sqrt(12.8 m/s)^2 +(3.6 m/s)^2`
`=13.3 m/s`.
The velocity makes an angle theta with the X-axis where
`tantheta= (v_y)/(v_x) = (3.6 m/s)/(12.8 m/s) = 9/32`
The x-coordiN/Ate at t=4.0 s is
` x= u_x t+1/2 a_x t^2`
`=(8.0 m/s)(4.0 s)+1/2 (1.2m/s^2)(4.0s)^2`
`=32m+9.6 m 41.6m.
The y=component at t=4.0 is
`y=u_y t+1/2 at^2`
`=1/2 (0.90 m/s^2)(4.0s)^2`
`=7.2m`
Thus, the particle is at (41.6 m, 7.2 m) at 4.0 s.