Question

A dog running in an open field has components of velocity vₓ = 2.8 m/s and Vᵧ = -1.4 m/s at time t₁ = 12.0 s. For the time interval from t₁ = 120 s to t₂ = 25.0 s, the average acceleration of the dog has magnitude 0.50 m/s² and direction 360° measured from the ti-axis toward the +y-axis. Note: We are looking down on the dog from above, so the y direction here is also on the field; you do not have to include gravity in this problem.

A. At time t₂ = 25.0 s, what is the 2-component of the dog's velocity? Express your answer to three significant figures with the appropriate units.

Answer 1

The y-component of the dog's velocity at t2 = 25.0 s is 5.1 m/s, calculated by using the initial velocity and the average acceleration over the given period.

Step-by-step explanation:

To determine the y-component of the dog's velocity at time t2 = 25.0 seconds, we need to use the following kinematic equation:

v = v0 + at

where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time elapsed.

Since the question already provides us with an average acceleration a = 0.50 m/s2 directed at 360° from the ti-axis towards the +y-axis, we can calculate the change in the y-component of the velocity.

The time interval Δt is t2 - t1 = 25.0 s - 12.0 s = 13.0 s. We're given that Vy1 = -1.4 m/s, therefore:

Vy2 = Vy1 + aΔt = -1.4 m/s + (0.50 m/s2)(13.0 s)

Vy2 = -1.4 m/s + 6.5 m/s = 5.1 m/s

So, the y-component of the dog's velocity at time t2 is 5.1 m/s.