How to find the components, magnitude, and direction of velocity?

Question

A dog running in an open field has velocity components of vx= 2.0 m/s and vy = -1.8 m/s at time t1=12.0s. For the time interval t1 = 12.0 s to t2 = 23.0 s, the average acceleration of the dog has magnitude 0.55m/s2 and direction 36.0 degrees from the +x-axis towards the +y-axis.

a. At time t2 = 23.0 s, what is the x-component of the dog's velocity?

b. At time t2 = 23.0 s, what is the y-component of the dog's velocity?

c. What is the magnitude of the dog's velocity?

d. What is the direction of the dog's velocity (measured from the +x-axis toward the +y-axis)?

Answer 1

Let

a = 0.55 m/s² be the magnitude of the acceleration,

α = 36° be the angle of the acceleration,

v_x(t) be the x-component of velocity at time t,

v_y(t) be the y-component of velocity at time t.

We are given

v_x(t₁) = 2 m/s,

v_y(t₁) = -1.8 m/s.

The component of the acceleration in the x-direction is a_x = acos(α).

The component of the acceleration in the y-direction is a_y = asin(α).

We apply the kinematic equations in each direction separately.

a.

We are to calculate v_x(t₂).

v_x(t₂) = v_x(t₁) + a_x(t₂ - t₁) =

v_x(t₁) + acos(α)(t₂ - t₁) =

2 + 0.55×cos(36°)×(23 - 12) = 6.9 m/s

b.

We are to calculate v_y(t₂).

v_y(t₂) = v_y(t₁) + a_y(t₂ - t₁) =

v_y(t₁) + asin(α)(t₂ - t₁) =

-1.8 + 0.55×sin(36°)×(23 - 12) = 1.8 m/s

c.

The magnitude of the velocity is

√(v_x(t₂)² + v_y(t₂)²) =

√(6.9² + 1.8²) = 7.1 m/s

d.

Let θ be the angle of the velocity at time t₂ against the x-axis.

Then

tan(θ) = v_y(t₂)/v_x(t₂)

θ = arctan(v_y(t₂)/v_x(t₂)) = arctan(1.8/6.9) = 14.6°.