Question

A dog running in an open field has components of velocity vx = 2.6 m/s and vy = -1.6 m/s at time t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 21.0 s, the average acceleration of the dog has magnitude 0.55 m/s² and direction 36.0° measured from the +x-axis toward the +y-axis. At time t = 21.0 s, what is the x-component of the dog's velocity? At time t = 21.0 s, what is the y-component of the dog's velocity? What is the magnitude of the dog's velocity? 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

The x-component of the dog's velocity at t = 21.0 s is 8.1 m/s. The y-component of the dog's velocity at t = 21.0 s is 3.9 m/s. The magnitude of the dog's velocity at t = 21.0 s is 8.99 m/s and the direction is 27.0° measured from the +x-axis toward the +y-axis.

Step-by-step explanation:

To find the x-component of the dog's velocity at time t = 21.0 s, we can use the formula:

vx = vx0 + ax * Δt

Substituting the given values, we have vx = 2.6 m/s + (0.55 m/s²) * (21.0 s - 10.0 s) = 2.6 m/s + 5.5 m/s = 8.1 m/s.

Similarly, to find the y-component of the dog's velocity at time t = 21.0 s, we can use the formula:

vy = vy0 + ay * Δt

Substituting the given values, we have vy = -1.6 m/s + (0.55 m/s²) * (21.0 s - 10.0 s) = -1.6 m/s + 5.5 m/s = 3.9 m/s.

To find the magnitude of the dog's velocity, we can use the formula:

v = sqrt(vx² + vy²)

Substituting the values, we have v = sqrt((8.1 m/s)² + (3.9 m/s)²) = sqrt(65.61 m²/s² + 15.21 m²/s²) = sqrt(80.82 m²/s²) = 8.99 m/s.

To find the direction of the dog's velocity, we can use the formula:

θ = arctan(vy / vx)

Substituting the values, we have θ = arctan(3.9 m/s / 8.1 m/s) = 0.47 radians = 27.0°.

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