A dog running in an open field has components of velocity vx = 2.8 m/s and vy = -1.2 m/s at time t1 = 12.0 s. For the time interval from t1 = 12.0 s to t2 = 22.0 s, the average acceleration of the dog has magnitude 0.50 m/s2 and direction 36.0 ∘ measured from the +x-axis toward the +y-axis.

Question

A dog running in an open field has components of velocity vx = 2.8 m/s and vy = -1.2 m/s at time t1 = 12.0 s. For the time interval from t1 = 12.0 s to t2 = 22.0 s, the average acceleration of the dog has magnitude 0.50 m/s2 and direction 36.0 ∘ measured from the +x-axis toward the +y-axis.

Part A
At time t2 = 22.0 s , what is the x -component of the dog's velocity?
Express your answer with the appropriate units.
Part B
At time t2 = 22.0 s , what is the y -component of the dog's velocity?
Express your answer with the appropriate units.
Part C
What is the magnitude of the dog's velocity?
Express your answer with the appropriate units.
Part D
What is the direction of the dog's velocity (measured from the +x -axis toward the +y -axis)?
Express your answer in degrees.

Answer 1

Do an x problem and a y problem

x problem
at t = 12 Vxi = 2.8 m/s
Ax = 0.5 cos 36 = 0.405 m/s^2
so
Vx = 2.8 + 0.405 (t-12)
so at t = 22
Vx = 2.8 + 0.405 (22-12) = 6.85 m/s (answer Part A)
Get it ? Now do the Y problem the same way.
I think you can do sqrt(Vx^2+Vy^2) etc

Answer 2

Part A: The x-component of velocity remains the same throughout the interval, so the x-component of the dog's velocity at t2 = 22.0 s is 2.8 m/s.

Part B: To find the y-component of velocity at t2 = 22.0 s, we need to use the average acceleration and time interval t2 - t1 = 22.0 s - 12.0 s = 10.0 s.

Using the formula vy = vyi + ayt, where vyi is the initial y-component of velocity, a is the acceleration, and t is the time, we have:
vy = (-1.2 m/s) + (0.50 m/s^2)(10.0 s) = -1.2 m/s + 5.0 m/s = 3.8 m/s

So the y-component of the dog's velocity at t2 = 22.0 s is 3.8 m/s.

Part C: The magnitude of velocity can be found using the Pythagorean theorem:
|v| = sqrt(vx^2 + vy^2)
= sqrt((2.8 m/s)^2 + (3.8 m/s)^2)
= sqrt(7.84 m^2/s^2 + 14.44 m^2/s^2)
= sqrt(22.28 m^2/s^2)
≈ 4.72 m/s

So the magnitude of the dog's velocity is approximately 4.72 m/s.

Part D: The direction of velocity can be found using the inverse tangent function:
θ = arctan(vy/vx)
= arctan((3.8 m/s)/(2.8 m/s))
≈ arctan(1.357)

Using a calculator, we find that arctan(1.357) ≈ 51.5 degrees.

So the direction of the dog's velocity, measured from the +x-axis toward the +y-axis, is approximately 51.5 degrees.

Answer 3

To solve this problem, we can use the equations for average velocity and average acceleration:

Average velocity:

v_avg = (v_f + v_i) / 2

Average acceleration:

a_avg = (v_f - v_i) / (t_f - t_i)

Part A:

We are given the x-component of the velocity (vx) at time t1 = 12.0s. To find the x-component of the velocity at time t2 = 22.0s, we can use the average acceleration and the equation for average velocity.

v_avg_x = (v_f_x + v_i_x) / 2

v_avg_x = (v_f_x + vx) / 2

a_avg = (v_f_x - vx) / (t_f - t1)

Rearranging the equation for v_f_x, we have:

v_f_x = vx + a_avg * (t_f - t1)

v_f_x = 2.8 m/s + 0.50 m/s^2 * (22.0s - 12.0s)

v_f_x = 2.8 m/s + 0.50 m/s^2 * 10.0s

v_f_x = 2.8 m/s + 5 m/s

v_f_x = 7.8 m/s

Therefore, the x-component of the dog's velocity at time t2 = 22.0s is 7.8 m/s.

Part B:

We are given the y-component of the velocity (vy) at time t1 = 12.0s. To find the y-component of the velocity at time t2 = 22.0s, we can use the average acceleration and the equation for average velocity.

v_avg_y = (v_f_y + v_i_y) / 2

v_avg_y = (v_f_y + vy) / 2

a_avg = (v_f_y - vy) / (t_f - t1)

Rearranging the equation for v_f_y, we have:

v_f_y = vy + a_avg * (t_f - t1)

v_f_y = -1.2 m/s + 0.50 m/s^2 * (22.0s - 12.0s)

v_f_y = -1.2 m/s + 0.50 m/s^2 * 10.0s

v_f_y = -1.2 m/s + 5 m/s

v_f_y = 3.8 m/s

Therefore, the y-component of the dog's velocity at time t2 = 22.0s is 3.8 m/s.

Part C:

The magnitude of the dog's velocity can be found using the Pythagorean theorem:

|v| = sqrt(vx^2 + vy^2)

|v| = sqrt((2.8 m/s)^2 + (-1.2 m/s)^2)

|v| = sqrt(7.84 m^2/s^2 + 1.44 m^2/s^2)

|v| = sqrt(9.28 m^2/s^2)

|v| = 3.04 m/s

Therefore, the magnitude of the dog's velocity at time t2 = 22.0s is 3.04 m/s.

Part D:

The direction of the dog's velocity can be found using the arctan function:

θ = tan^(-1)(vy / vx)

θ = tan^(-1)(-1.2 m/s / 2.8 m/s)

θ = tan^(-1)(-0.4286)

θ = -23.45°

Since the direction is measured from the +x-axis toward the +y-axis, we need to add 90° to the angle:

θ = -23.45° + 90°

θ ≈ 66.55°

Therefore, the direction of the dog's velocity (measured from the +x-axis toward the +y-axis) at time t2 = 22.0s is approximately 66.55°.

Answer 4

To solve this problem, we first need to understand the given information and then apply the appropriate equations to find the answers. Let's break it down step by step:

Given information:
- The velocity components at time t1 = 12.0 s: vx = 2.8 m/s (in the x-direction) and vy = -1.2 m/s (in the y-direction).
- The time interval from t1 = 12.0 s to t2 = 22.0 s.
- The average acceleration during this interval: magnitude = 0.50 m/s^2 and direction = 36.0° measured from the +x-axis toward the +y-axis.

Part A:
To find the x-component of the dog's velocity at time t2 = 22.0 s, we can use the equation:
vx final = vx initial + ax * Δt

Here, vx initial is the x-component of the dog's velocity at t1.
ax is the x-component of the average acceleration, which can be found using trigonometry: ax = average acceleration * cos(angle).
Δt is the time interval, which is t2 - t1.

Applying the equation:
vx final = 2.8 m/s + (0.50 m/s^2 * cos(36.0°)) * (22.0 s - 12.0 s)

Solving this equation will give us the x-component of the dog's velocity at t2.

Part B:
To find the y-component of the dog's velocity at time t2 = 22.0 s, we can use the equation:
vy final = vy initial + ay * Δt

Here, vy initial is the y-component of the dog's velocity at t1.
ay is the y-component of the average acceleration, which can be found using trigonometry: ay = average acceleration * sin(angle).
Δt is the time interval, which is t2 - t1.

Applying the equation:
vy final = -1.2 m/s + (0.50 m/s^2 * sin(36.0°)) * (22.0 s - 12.0 s)

Solving this equation will give us the y-component of the dog's velocity at t2.

Part C:
The magnitude of the dog's velocity at time t2 can be found using the Pythagorean theorem:
velocity magnitude = sqrt((vx)^2 + (vy)^2)

Using the x-component and y-component of the dog's velocity obtained in Part A and Part B respectively, we can calculate the magnitude of the dog's velocity at t2.

Part D:
The direction of the dog's velocity is measured from the +x-axis toward the +y-axis. We can find this direction using the inverse tangent function:
velocity direction = atan(vy/vx)

Using the x-component and y-component of the dog's velocity obtained in Part A and Part B respectively, we can calculate the direction of the dog's velocity at t2.

Now you have all the information and equations needed to solve this problem!