Question

A dog running in an open field has components of velocity vₓ= 2.7 m/s and vᵧ = -1.1 m/s at time t1 = 11.2 s . For the time interval from t1 = 11.2 s to t2 = 24.0 s , the average acceleration of the dog has magnitude 0.49 m/s² and direction 26.5° measured from the +x-axis toward the +y-axis.

(a) At time t2 = 24.0 s, what are the x- and y-components of the dog's velocity?
(b) What is the magnitude of the dog's velocity?
(c) What is the direction of the dog's velocity (measured from the toward the + x-axis and +y-axis)?

Answer 1

a) vₓ = 8.3128 m / s , v_{y} = 1,698 m / s , b)v = 8.484 m / s , c) θ = 83.14°

Step-by-step explanation:

Let's use the kinematic equations for two-dimensional movement

Let's look for the components of acceleration using trigonometry

sin26.5 = ay / a

cos 26..5 = ax / a

ay = a sin26.5

ax = a cos 26.5

ay = 0.49 sin26.5

ax = 0.49 cos 26.5

ay = 0.2186 m / s2

ax = 0.4385 m / s2

a) Having the accelerations we look for the speed in each axis

vₓ = v₀ₓ + ax (t-t₀)

vₓ = 2.7 + 0.4385 (24.0 - 11.2)

vₓ = 8.3128 m / s

`v_(y)` = v_{oy} + a_{y} (t-to)

v_{y} = -1.1 + 0.2186 (24.0 - 11.2)

b) let's use Pythagoras' theorem to find the magnitude

v = √ Vₓ² + v_{y}²

v = Ra (8.3128² + 1,698²)

v = 8.484 m / s

c) Let's use trigonometry to find the angles

tan tea = v_{y} / vₓ

θ = tan⁻¹ v_{y} / vₓ

θ = tan⁻¹ 1,698 / 8.3128

θ = 83.14°