Aparticle 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

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

the body's situation won't change the forces are equal opposit and on the same body