speed and velocity

Question

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 8.5 m, y = 5 m, and has velocity ~vo = (4.5 m/s) ˆı + (−2 m/s) ˆ . The acceleration is given by ~a = (8 m/s 2 ) ˆı + (3.5 m/s 2 ) ˆ . What is the x component of velocity after 1.5 s?

a) Answer in units of m/s.

What is the y component of velocity after 1.5 s?

b) Answer in units of m/s.

What is the magnitude of the displacement from the origin (x = 0 m, y = 0 m) after 1.5 s?

c) Answer in units of m.

Answer 1

Assuming the Vo vector is 4.5i -2j m/s and the acceleration vector is 8i + 3.5j m/s² then:

For a) v_x = v_xi + a_xt = 4.5 + 8(1.5) = 16.5 m/s. The problem actually only has 1 significant figure in the numbers given so, if you maintain 1 sig fig, then it rounds the answer to 20 m/s. I'm unsure if you are doing that in your class so I'll leave the other answers in full mathematical form and let you round based on your teacher's requirements.

For b) v_y = v_yi + a_yt = -2 + 3.5(1.5) = 3.25 m/s

For c):

s_x = s_xi + v_xit + 1/2a_xt² = 8.5 + 4.5(1.5) + 1/2(8)(1.5²) = 24.25 m (this is the x component of the displacement vector)

s_y = s_yi + v_yit + 1/2a_yt² = 5 + (-2)(1.5) +1/2(3.5)(1.5²) = 5.9375 m (this is the y component of the displacement vector)

||s|| = sqrt(s_x² + s_y²) = sqrt(24.25² + 5.9375²) = 24.966 m (total magnitude of the displacement)