A snowmobile is originally at the point with position vector 28.7 m at 95.0° counterclockwise from the x axis, moving with velocity 4.85 m/s at 40.0°. It moves with constant acceleration 1.94 m/s2 at 200°. After 5.00 s have elapsed, find the following:

(a)Its velocity vector
(b)Its position vector

s0 = 28.7 @ 95.0° = -2.501,28.591

v0 = 4.85 @ 40.0° = 3.715,3.119
a = 1.94 @ 200° = -1.823,-0.664

Now, we just plug in our usual equations:

v = v0 + at
s = s0 + v0*t + 1/2 at^2

and then convert back to polar form as needed.

To find the velocity vector and position vector of the snowmobile after 5.00 seconds, we need to calculate both the magnitude and direction of these vectors at that time.

First, let's find the final velocity vector of the snowmobile. We can decompose the initial velocity vector given by 4.85 m/s at 40.0° into its x and y components using the sine and cosine functions:

Vx = 4.85 m/s * cos(40.0°) = 3.716 m/s
Vy = 4.85 m/s * sin(40.0°) = 3.114 m/s

Next, we can calculate the change in velocity, Δv, using the constant acceleration:

Δv = acceleration * time = 1.94 m/s² * 5.00 s = 9.7 m/s

Then, find the components of Δv in the x and y directions using the sine and cosine functions:

Δvx = Δv * cos(200°) = 9.7 m/s * cos(200°) = -9.517 m/s
Δvy = Δv * sin(200°) = 9.7 m/s * sin(200°) = -3.785 m/s

To find the final velocity vector, add the components of the initial velocity vector and the change in velocity:

Vxf = Vx + Δvx = 3.716 m/s + (-9.517 m/s) = -5.801 m/s
Vyf = Vy + Δvy = 3.114 m/s + (-3.785 m/s) = -0.671 m/s

The magnitude of the final velocity vector is given by the square root of the sum of the squares of its components:

|Vf| = √((-5.801 m/s)² + (-0.671 m/s)²) = 5.837 m/s

The direction of the final velocity vector can be found using the inverse tangent function:

θf = tan^(-1)(Vyf / Vxf) = tan^(-1)(-0.671 m/s / -5.801 m/s) = 6.897°

Therefore, the final velocity vector after 5.00 seconds is approximately 5.837 m/s at an angle of 6.897° counterclockwise from the x-axis.

Now, let's find the final position vector of the snowmobile.

The velocity vector can be used to find the displacement (change in position) of the snowmobile. The displacement is given by the product of the velocity and time:

Δx = Vx * t = 3.716 m/s * 5.00 s = 18.58 m
Δy = Vy * t = 3.114 m/s * 5.00 s = 15.57 m

To find the final position vector, add the components of the initial position vector (28.7 m at 95.0° counterclockwise from the x-axis) and the displacement:

xf = xi + Δx = 28.7 m + 18.58 m = 47.28 m
yf = yi + Δy = 0 m + 15.57 m = 15.57 m

Therefore, the final position vector of the snowmobile after 5.00 seconds is approximately (47.28 m, 15.57 m).