Question

The acceleration of a particle is a constant. At t = 0 the velocity of the particle is (10ˆi+20ˆj)m/s. At t = 4 s the velocity is 10ˆjm/s. (a) What is the particle’s acceleration? (b) How do the position and velocity vary with time? Assume the particle is initially at the origin.

a) (i) -2.5i-5.0j m/s²
(ii) -5.0i-10.0j m/s²
(iii) -2.5i-10.0j m/s²
(iv) -5.0i-5.0j m/s²

b) (i) r(t)=10tˆi+20tˆj-5t²ˆi-10t²ˆj m, v(t)=10ˆi+20ˆj-10tˆi-20tˆj m/s
(ii) r(t)=10tˆi+20tˆj-5t²ˆi-10t²ˆj m, v(t)=10ˆi+20ˆj-5tˆi-10tˆj m/s
(iii) r(t)=10tˆi+20tˆj-5t²ˆi-10t²ˆj m, v(t)=10ˆi+20ˆj-5tˆi-20tˆj m/s
(iv) r(t)=10tˆi+20tˆj-5t²ˆi-10t²ˆj m, v(t)=10ˆi+20ˆj-10tˆi-10tˆj m/s

Answer 1

The acceleration of the particle is -2.5i - 2.5j m/s². The position and velocity functions are given by r(t) = (10t - 1.25t²)i + (20t - 1.25t²)j m and v(t) = (10 - 2.5t)i + (20 - 2.5t)j m/s respectively.

Step-by-step explanation:

The problem pertains to a particle moving with constant acceleration. The acceleration can be found by examining the change in velocity over time. The velocity at t = 0 is given as (10⁰i + 20⁰j) m/s, and at t = 4 s, it is 10⁰j m/s. The acceleration components in i and j can be calculated as follows:

a_i = (v_{4s, i} - v_{0s, i}) / Δt = (0 - 10) m/s / 4 s = -2.5 m/s²

a_j = (v_{4s, j} - v_{0s, j}) / Δt = (10 - 20) m/s / 4 s = -2.5 m/s²

Thus, the particle's acceleration is -2.5i - 2.5j m/s², which matches option (iii).

For part (b), the position and velocity as functions of time can be determined using the equations of motion for constant acceleration:

• v(t) = v_0 + a*t
• r(t) = r_0 + v_0*t + (1/2)*a*t²

Since the particle starts at the origin with initial velocity (10⁰i + 20⁰j) m/s and acceleration -2.5⁰i - 2.5⁰j m/s², we get:

v(t) = (10 - 2.5t)⁰i + (20 - 2.5t)⁰j m/s

r(t) = (10t - 1.25t²)⁰i + (20t - 1.25t²)⁰j m

These functions show the variation in the particle's position and velocity over time, with option (iii) being the correct one.