A dog running in an open field has components of velocity vx = 2.5 m/s and vy = -1.3 m/s at time t1 = 10.7 s . For the time interval from t1 = 10.7 s to t2 = 23.2 s , the average acceleration of the dog has magnitude 0.54 m/s2 and direction 25.0 ∘ measured from the +x−axis toward the +y−axis.

Question

A dog running in an open field has components of velocity vx = 2.5 m/s and vy = -1.3 m/s at time t1 = 10.7 s . For the time interval from t1 = 10.7 s to t2 = 23.2 s , the average acceleration of the dog has magnitude 0.54 m/s2 and direction 25.0 ∘ measured from the +x−axis toward the +y−axis.

Answer 1

One wonders what the question is.

Answer 2

To find the average acceleration of the dog during the given time interval, we can use the following formula:

average acceleration = (change in velocity) / (time interval)

First, let's find the change in velocity. We are given the components of the initial velocity (vx = 2.5 m/s, vy = -1.3 m/s) at time t1 = 10.7 s, and we need to find the final components of velocity at time t2 = 23.2 s.

To find the final velocity components, we need to consider the average acceleration and the time interval:

average acceleration = magnitude = 0.54 m/s^2
direction = 25.0° measured from the +x-axis toward the +y-axis

We can use the following equations to find the final components of velocity:

vxf = vx0 + (ax * t)
vyf = vy0 + (ay * t)

where:
vxf and vyf are the final velocity components (to be solved)
vx0 = 2.5 m/s (initial vx component)
vy0 = -1.3 m/s (initial vy component)
ax = average acceleration * cos(direction)
ay = average acceleration * sin(direction)
t = time interval = t2 - t1 = 23.2 s - 10.7 s = 12.5 s

Let's plug in the values:

ax = 0.54 m/s^2 * cos(25°)
ay = 0.54 m/s^2 * sin(25°)
t = 12.5 s

ax ≈ 0.487 m/s^2
ay ≈ 0.229 m/s^2

Now we can calculate the final velocity components:

vxf = vx0 + (ax * t)
= 2.5 m/s + (0.487 m/s^2 * 12.5 s)

vyf = vy0 + (ay * t)
= -1.3 m/s + (0.229 m/s^2 * 12.5 s)

By plugging in the values and performing the calculations:

vxf ≈ 2.5 m/s + 6.0875 m/s = 8.5875 m/s
vyf ≈ -1.3 m/s + 2.8625 m/s = 1.5625 m/s

So, the final velocity components at time t2 = 23.2 s are vxf ≈ 8.5875 m/s and vyf ≈ 1.5625 m/s.

Now that we have the initial and final velocity components, we can find the change in velocity:

Δvx = vxf - vx0 = 8.5875 m/s - 2.5 m/s = 6.0875 m/s
Δvy = vyf - vy0 = 1.5625 m/s - (-1.3 m/s) = 2.8625 m/s

Finally, we can calculate the average acceleration using the formula mentioned earlier:

average acceleration = (change in velocity) / (time interval)
= √(Δvx^2 + Δvy^2) / t
= √((6.0875 m/s)^2 + (2.8625 m/s)^2) / 12.5 s
≈ √(37.18 m^2/s^2 + 8.191 m^2/s^2) / 12.5 s
≈ √(45.371 m^2/s^2) / 12.5 s
≈ 6.741 m/s^2 / 12.5 s
≈ 0.539 m/s^2

Please note that due to rounding errors, the magnitude of the average acceleration might not be exactly 0.54 m/s^2, but it should be close.

So, the average acceleration of the dog during the time interval from t1 = 10.7 s to t2 = 23.2 s is approximately 0.54 m/s^2, as given.