A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration 20.320 m/s2. Suppose it moves as a particle under constant acceleration for 4.50 s. Find (a) its position and (b) its velocity at the end of this time interval. Next, assume it moves as a particle in simple harmonic motion for 4.50 s and x 5 0 is its equilibrium position. Find (c) its position and (d) its velocity at the end of this time interval.

a) - 2.34 m

b) - 1.3 m/s

c) 0.056 m

d) 0.320 m/s

Explanation:

part a

Given:

s (0) = 0.27 m

v (0) = 0.14 m/s

a = - 0.320 m/s^2

t = 4.50s

Using kinematic equation of motion for constant acceleration:

s (t) = s (0) + v (0) * t + 0.5*a*t^2

s (4.5 s) = 0.27 + 0.14*4.5 + 0.5*-0.32*4.5^2

s (4.5) = - 2.34 m

part b

Using kinematic equation of motion for constant acceleration:

v (t) = v (0) + a*t

v (4.5s) = (0.14) + (-0.32) (4.5) = - 1.3 m/s

part c

Use equation for simple harmonic motion:

s (t) = A*cos (w*t)

v (t) = - A*w*sin (w*t)

a (t) = - A * (w) ^2 * cos (w*t)

0.27 m = A*cos (w*t) ... Eq 1

0.14 m/s = - A*w*sin (w*t) ... Eq2

-0.320 = - A * (w) ^2 * cos (w*t) ... Eq3

Solve the three equations above for A, w, and t

Divide 1 and 3:

w^2 = 0.32 / 0.27

w = 1.0887 rad / s

Divide 2 and 1:

w*tan (wt) = 0.14 / 0.27

tan (1.0887*t) = 0.476289

t = 0.4083 s

A = 0.27 / cos (1.0887*0.4083) = 0.3 m

Hence, the SHM is s (t) = 0.3*cos (1.0887*t)

s (4.5) = 0.3*cos (1.0887*4.5) = 0.056 m

part d

v (t) = - 0.32661 * sin (1.0887*t)

v (4.5) = - 0.32661 * sin (1.0887*4.5) = 0.320 m/s