Problem 1- A dog running in an open field has components of velocity vx = 2.2m/s and vy = -2.5m/s at time t1 = 12 s. For the time interval from t1 = 12 s to t2 = 24 s , the average acceleration of the dog has magnitude 0.55 m/s2 and direction 25.5∘ measured from the +x−axis counterclockwise. At t2 = 24 s What is the speed of the dog and what is the direction of its motion?
A dog running in an open field has components of velocity vx = 3.2 m/s and vy = -1.6 m/s at time t1 = 11.0 s . For the time interval from t1 = 11.0 s to t2 = 25.0 s , the average acceleration of the dog has magnitude 0.31 m/s2 and direction 36.0 ∘ measured from the +x−axis toward the +y−axis.
To find the speed of the dog at t2 = 24 s, we can use the average acceleration and the time interval.
Step 1: Calculate the change in velocity in the x-direction.
Acceleration in the x-direction (ax) = acceleration * cos(angle)
ax = 0.55 m/s^2 * cos(25.5°)
Step 2: Calculate the change in velocity in the y-direction.
Acceleration in the y-direction (ay) = acceleration * sin(angle)
ay = 0.55 m/s^2 * sin(25.5°)
Step 3: Calculate the change in velocity in the x-direction using the time interval.
Change in velocity in the x-direction (Δvx) = ax * Δt
Δvx = ax * (t2 - t1)
Step 4: Calculate the change in velocity in the y-direction using the time interval.
Change in velocity in the y-direction (Δvy) = ay * Δt
Δvy = ay * (t2 - t1)
Step 5: Calculate the speed of the dog at t2.
Speed (V) = √((vx + Δvx)^2 + (vy + Δvy)^2)
Step 6: Calculate the direction of the dog's motion at t2.
Direction (θ) = tan^(-1)((vy + Δvy) / (vx + Δvx))
Now, let's substitute the given values and calculate the required values.
Given:
vx = 2.2 m/s
vy = -2.5 m/s
t1 = 12 s
t2 = 24 s
acceleration = 0.55 m/s^2
angle = 25.5°
Step 1:
ax = 0.55 m/s^2 * cos(25.5°)
ax ≈ 0.4907 m/s^2
Step 2:
ay = 0.55 m/s^2 * sin(25.5°)
ay ≈ 0.233 m/s^2
Step 3:
Δvx = ax * (t2 - t1)
Δvx = 0.4907 m/s^2 * (24 s - 12 s)
Δvx = 5.8884 m/s
Step 4:
Δvy = ay * (t2 - t1)
Δvy = 0.233 m/s^2 * (24 s - 12 s)
Δvy = 2.796 m/s
Step 5:
Speed (V) = √((vx + Δvx)^2 + (vy + Δvy)^2)
Speed (V) = √((2.2 m/s + 5.8884 m/s)^2 + (-2.5 m/s + 2.796 m/s)^2)
Speed (V) ≈ √(8.0884 m/s)^2 + (0.296 m/s)^2)
Speed (V) ≈ √(65.4153 m^2/s^2 + 0.0875 m^2/s^2)
Speed (V) ≈ √(65.5028 m^2/s^2)
Speed (V) ≈ 8.107 m/s
Step 6:
Direction (θ) = tan^(-1)((vy + Δvy) / (vx + Δvx))
Direction (θ) = tan^(-1)((-2.5 m/s + 2.796 m/s) / (2.2 m/s + 5.8884 m/s))
Direction (θ) = tan^(-1)(-0.266 m/s / 8.0884 m/s)
Direction (θ) ≈ tan^(-1)(-0.0329)
Direction (θ) ≈ -1.88°
Therefore, the speed of the dog at t2 = 24 s is approximately 8.107 m/s, and the direction of its motion is approximately -1.88° (counterclockwise from the +x-axis).
To find the speed of the dog at time t2 = 24 s and the direction of its motion, we need to calculate the magnitude and direction of its velocity at t2.
First, let's find the components of the dog's acceleration. We know that the average acceleration has a magnitude of 0.55 m/s^2 and a direction of 25.5 degrees measured counterclockwise from the +x-axis.
We can break down the acceleration into its x and y components using trigonometry. The x-component of acceleration (ax) is given by:
ax = a * cos(theta)
where a is the magnitude of the acceleration and theta is the direction of the acceleration.
Plugging in the given values, we have:
ax = 0.55 m/s^2 * cos(25.5 degrees)
Next, we find the y-component of acceleration (ay):
ay = a * sin(theta)
where a is the magnitude of the acceleration and theta is the direction of the acceleration.
Plugging in the given values, we have:
ay = 0.55 m/s^2 * sin(25.5 degrees)
Now let's find the change in time:
delta t = t2 - t1 = 24 s - 12 s = 12 s
To calculate the change in velocity, we use the following equations:
delta vx = ax * delta t
delta vy = ay * delta t
Using the values we found earlier, we have:
delta vx = (ax) * (delta t) = (0.55 m/s^2 * cos(25.5 degrees)) * (12 s)
delta vy = (ay) * (delta t) = (0.55 m/s^2 * sin(25.5 degrees)) * (12 s)
Next, we find the final velocity by adding the change in velocity to the initial velocity at t1:
vx2 = vx1 + delta vx
vy2 = vy1 + delta vy
Plugging in the given values, we have:
vx2 = 2.2 m/s + (0.55 m/s^2 * cos(25.5 degrees)) * (12 s)
vy2 = -2.5 m/s + (0.55 m/s^2 * sin(25.5 degrees)) * (12 s)
To find the speed of the dog at t2, we use the Pythagorean theorem:
speed (v2) = sqrt((vx2)^2 + (vy2)^2)
Plugging in the values we found earlier, we have:
speed (v2) = sqrt((2.2 m/s + (0.55 m/s^2 * cos(25.5 degrees)) * (12 s))^2 + (-2.5 m/s + (0.55 m/s^2 * sin(25.5 degrees)) * (12 s))^2)
Finally, to find the direction of the dog's motion, we use trigonometry again. The direction (theta2) can be calculated using the arc tangent function:
theta2 = atan(vy2 / vx2)
Plugging in the values we found earlier, we have:
theta2 = atan((-2.5 m/s + (0.55 m/s^2 * sin(25.5 degrees)) * (12 s)) / (2.2 m/s + (0.55 m/s^2 * cos(25.5 degrees)) * (12 s)))
Solving these equations will give you the speed of the dog at t2 and the direction of its motion.