Question

At t= 0 s, a ball is moving with constant deceleration has a velocity vector with magnitude 3.61 m/s and a direction of 33.69o south of east. At t=3.00 s, the same ball has a velocity vector with a magnitude of 11.40 m/s and a direction of 37.88o north of of east. Find the magnitude of the acceleration of the ball.

Answer 1

3.61 m/s²

Step-by-step explanation:

Find the x and y components of the initial and final velocities.

uₓ = 3.61 cos(-33.69) = 3.00 m/s

uᵧ = 3.61 sin(-33.69) = -2.00 m/s

vₓ = 11.40 cos(37.88) = 9.00 m/s

vᵧ = 11.40 sin(37.88) = 7.00 m/s

Find the x and y components of the acceleration.

aₓ = (vₓ − uₓ) / t = (9.00 − 3.00) / 3.00 = 2.00 m/s²

aᵧ = (vᵧ − uᵧ) / t = (7.00 − -2.00) / 3.00 = 3.00 m/s²

Use Pythagorean theorem to find the magnitude.

a² = aₓ² + aᵧ²

a² = (2.00)² + (3.00)²

a = 3.61 m/s²