Question

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.

Answer 1

Answer:

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

Answer 2

The position and velocity of the particle in equilibrium position at the given parameters are;

A) x = -2.34 m

B) v_x = -1.3 m/s

C) x(4.5) = -0.076 m

D) v = 0.314 m/s

We are given;

Initial distance; x_i = 0.27 m

Initial velocity; v_xi = 0.14 m/s

Acceleration; a_x = - 0.32 m/s²

Time; t = 4.5 s

A) Formula for the particle's position as a function of time under constant acceleration is;

x = x_i + v_xi•t + ½a_x•t²

x = 0.27 + (0.14 × 4.5) + ½(-0.32 × 4.5²)

x = -2.34 m

B) Formula for it's velocity at the end of the time interval is;

v_x = v_xi + a_x•t

v_x = 0.14 + (-0.32 × 4.5)

v_x = -1.3 m/s

C) Formula for position in simple harmonic motion is;

x(t) = A cos(ωt + ϕ)

We know that acceleration is;

a = -ω²x

Thus;

-0.32 = -ω²(0.27)

ω = √(0.32/0.27)

ω = 1.089 rad/s

Now, velocity is the derivative of x(t). Thus;

v(t) = x'(t) = -Aω sin (ωt + ϕ)

At t = 0, we have;

0.14 = -A(1.089) × sin ϕ  - - -(1)

Also, at t = 0,

0.27 = A cos ϕ  - - - (2)

Divide equation 1 by equation 2 to get;

0.14/0.27 = -1.089 tan ϕ

ϕ = tan^(-1) (0.14/(0.27 × -1.089))

ϕ = -25.46°

Thus, putting -25.46° for ϕ in eq 2 gives;

0.27 = A cos (-25.46)

0.27 = A × 0.90183

A = 0.27/0.90183

A = 0.2994

Thus,

x(4.5) = 0.2994 cos((1.089 × 4.5) + (-25.46))

x(4.5) = -0.076 m

D) v = -0.2994 × 1.089 × sin 254.6

v = 0.314 m/s

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