A snowmobile is originally at the point with position vector 26.1 m at 95.0° counterclockwise from the x axis, moving with velocity 4.25 m/s at 40.0°. It moves with constant acceleration 1.73 m/s2 at 200°. After 5.00 s have elapsed, find the velocity vector, and position vector

To find the velocity vector and position vector of the snowmobile after 5.00 seconds, we can use the equations of motion in two dimensions.

1. Finding the velocity vector:
The initial velocity vector is given by v0 = 4.25 m/s at 40.0°. Since the snowmobile is moving with constant acceleration, we can use the equation:

v = v0 + at,

where v is the final velocity vector, a is the acceleration vector, and t is the time elapsed.

First, let's break down the initial velocity vector into its x and y components:
v0x = v0 * cos(40.0°),
v0y = v0 * sin(40.0°).

Now, let's calculate the final velocity components:
vx = v0x + a * cos(200°) * t,
vy = v0y + a * sin(200°) * t.

Then, we can combine these components to get the final velocity vector:
v = √(vx^2 + vy^2) with direction tan^(-1)(vy / vx).

2. Finding the position vector:
The position vector is given by:
r = r0 + v0t + (1/2)at^2,

where r is the final position vector, r0 is the initial position vector, v0 is the initial velocity vector, a is the acceleration vector, and t is the time elapsed.

First, let's break down the initial position vector into its x and y components:
r0x = 26.1 m * cos(95.0°),
r0y = 26.1 m * sin(95.0°).

Now, let's calculate the final position components:
rx = r0x + v0x * t + (1/2) * a * cos(200°) * t^2,
ry = r0y + v0y * t + (1/2) * a * sin(200°) * t^2.

Then, we can combine these components to get the final position vector:
r = √(rx^2 + ry^2) with direction tan^(-1)(ry / rx).

Using these equations, you can calculate the velocity vector and position vector of the snowmobile after 5.00 seconds.