a Particle in x-y plane with const acceleration of 1.5m/s sq. in the direction making an angle 37degree with x_axis.

At t=0 the particle is at origin and its velocity is 8m/s along x axis.Find the velocity and the position of the particle at t=4sec.

V = 8 + 1.5[37o]*4 = 8 + 6[37o] = 8 + 6*Cos37+6*sin37 = 8 + 4.79+3.61i = 12.79 + 3.61i = 13.3m/s[15.8o]

To find the velocity and position of the particle at t=4 sec, we can break down the acceleration into its x and y components. Given that the acceleration is constant and making an angle of 37 degrees with the x-axis, we can find these components as:

ax = a * cos(θ)
ay = a * sin(θ)

where a is the magnitude of the acceleration (1.5 m/s^2) and θ is the angle (37 degrees).

Calculating the components:
ax = 1.5 m/s^2 * cos(37 degrees)
ay = 1.5 m/s^2 * sin(37 degrees)

Using these values, we can find the velocity components using the equations of motion:

vx = ux + ax * t
vy = uy + ay * t

where ux and uy are the initial velocities in the x and y directions, and t is the time (4 sec).

Given that the initial velocity in the x-direction is 8 m/s (ux = 8 m/s), we have:

vx = 8 m/s + ax * 4 s
vy = 0 m/s + ay * 4 s

Substituting the calculated values of ax, ay, and ux:

vx = 8 m/s + (1.5 m/s^2 * cos(37 degrees)) * 4 s
vy = 0 m/s + (1.5 m/s^2 * sin(37 degrees)) * 4 s

Now, let's calculate the position components using the following equations:

x = ux * t + (1/2) * ax * t^2
y = uy * t + (1/2) * ay * t^2

where x and y are the position coordinates.

Given that the initial position is at the origin (0,0), we have:

x = 0 + (1/2) * (1.5 m/s^2 * cos(37 degrees)) * (4 s)^2
y = 0 + (1/2) * (1.5 m/s^2 * sin(37 degrees)) * (4 s)^2

Now, let's calculate the final values:

vx = 8 m/s + (1.5 m/s^2 * cos(37 degrees)) * 4 s
vy = 0 m/s + (1.5 m/s^2 * sin(37 degrees)) * 4 s

x = (1/2) * (1.5 m/s^2 * cos(37 degrees)) * (4 s)^2
y = (1/2) * (1.5 m/s^2 * sin(37 degrees)) * (4 s)^2

Simplifying these expressions will give us the velocity and position of the particle at t=4 sec.