A snowmobile is originally at the point with position vector 28.7 m at 95.0° counterclockwise from the x axis, moving with velocity 4.13 m/s at 40.0°. It moves with constant acceleration 2.05 m/s2 at 200°. After 5.00 s have elapsed, find the following.
(a) its velocity vector
v =
(b) its position vector
r =
(x,y) at start:
x = 28.7 cos 95 = 28.7 * -0.0872 = -2.50
y = 28.7 sin 95 = +28.6
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velocity at t = 0
Vix = 4.13 cos 40 = 3.16
Viy = 4.13 sin 40 = 2.65
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ax = 2.05 cos 200 = -1.93
ay = 2.05 sin 200 = -0.701
velocity at t = 5
Vx = 3.16 -1.93 (5)= -6.49m/s
Vy = 2.65 -0.701(5) = -0.855 m/s
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X = -2.50 + 3.16 * 5 - (1/2)(1.93)(25)
Y = 28.6 +2.65*5 - (1/2)(0.701)(25)
(a) its velocity vector
v = 3.85 m/s at 82° counterclockwise from the x axis
(b) its position vector
r = 10.15 m at 135° counterclockwise from the x axis
To solve this problem, we will use vector addition to find the velocity and position vectors of the snowmobile.
(a) Finding the velocity vector:
1. Start by breaking down the initial velocity vector into its x and y components.
Given: magnitude = 4.13 m/s, angle = 40.0°
Vx = magnitude * cos(angle)
= 4.13 m/s * cos(40°)
≈ 3.16 m/s
Vy = magnitude * sin(angle)
= 4.13 m/s * sin(40°)
≈ 2.64 m/s
2. Next, calculate the change in velocity vector due to acceleration over 5.00 seconds.
Given: acceleration = 2.05 m/s^2, angle = 200°, time = 5.00 s
Change in velocity magnitude = acceleration * time
= 2.05 m/s^2 * 5.00 s
= 10.25 m/s
Change in velocity x-component = -change in velocity magnitude * cos(angle)
= -10.25 m/s * cos(200°)
≈ -7.82 m/s
Change in velocity y-component = -change in velocity magnitude * sin(angle)
= -10.25 m/s * sin(200°)
≈ -2.98 m/s
3. Add the initial velocity components and the change in velocity components together to find the final velocity vector.
Vx_final = Vx_initial + Change in Vx
= 3.16 m/s + (-7.82 m/s)
≈ -4.66 m/s
Vy_final = Vy_initial + Change in Vy
= 2.64 m/s + (-2.98 m/s)
≈ -0.34 m/s
V_final = √(Vx_final^2 + Vy_final^2)
= √((-4.66 m/s)^2 + (-0.34 m/s)^2)
≈ 4.67 m/s
θ_final = tan^(-1)(Vy_final / Vx_final)
= tan^(-1)((-0.34 m/s) / (-4.66 m/s))
≈ 4.64°
Therefore, the final velocity vector after 5.00 seconds is approximately v = 4.67 m/s at 4.64° counterclockwise from the x-axis.
(b) Finding the position vector:
1. Start by breaking down the initial position vector into its x and y components.
Given: magnitude = 28.7 m, angle = 95.0°
Rx = magnitude * cos(angle)
= 28.7 m * cos(95.0°)
≈ -4.72 m
Ry = magnitude * sin(angle)
= 28.7 m * sin(95.0°)
≈ 28.32 m
2. Next, calculate the change in position vector due to the initial velocity over 5.00 seconds.
Given: Vx_initial = 3.16 m/s, Vy_initial = 2.64 m/s, time = 5.00 s
Change in position x-component = Vx_initial * time
= 3.16 m/s * 5.00 s
= 15.8 m
Change in position y-component = Vy_initial * time
= 2.64 m/s * 5.00 s
= 13.2 m
3. Add the initial position components and the change in position components together to find the final position vector.
Rx_final = Rx_initial + Change in Rx
= -4.72 m + 15.8 m
≈ 11.08 m
Ry_final = Ry_initial + Change in Ry
= 28.32 m + 13.2 m
≈ 41.52 m
Therefore, the final position vector after 5.00 seconds is approximately r = 11.08 m at 41.52 m counterclockwise from the x-axis.
To find the answers to the given questions, we need to use vector addition and displacement formulas. Let's break down the problem step by step:
(a) To find the velocity vector, we need to find the resultant of the initial velocity and the acceleration vectors.
Given:
Initial velocity = 4.13 m/s at 40.0° counterclockwise from the x-axis
Acceleration = 2.05 m/s^2 at 200° counterclockwise from the x-axis
Convert the given angles into standard position angles:
40.0° counterclockwise from the x-axis = 360° - 40.0° = 320.0°
200° counterclockwise from the x-axis = 360° - 200° = 160.0°
Now, let's break down the initial velocity vector into its x and y components:
Vx = 4.13 m/s * cos(320.0°)
Vy = 4.13 m/s * sin(320.0°)
Calculate the acceleration vector components:
Ax = 2.05 m/s^2 * cos(160.0°)
Ay = 2.05 m/s^2 * sin(160.0°)
Now, add the x and y components separately to find the resultant velocity components:
Vx_resultant = Vx + Ax * t
Vy_resultant = Vy + Ay * t
Where t = 5.00 s (time elapsed)
Finally, combine the resultant velocity components to find the velocity vector:
v = √(Vx_resultant^2 + Vy_resultant^2) at an angle θ = tan^(-1)(Vy_resultant / Vx_resultant)
Substitute the calculated components and find the magnitude and angle of the resultant velocity vector.
(b) To find the position vector, we will use the displacement formula:
r = r0 + v0 * t + 0.5 * a * t^2
Given:
r0 = 28.7 m at 95.0° counterclockwise from the x-axis (initial position vector)
v0 = initial velocity vector (calculated in part a)
a = acceleration vector
Calculate the displacement vector by substituting the values into the displacement formula.
Now, let's calculate the answers using the given data.