A particle moves along the x axis. It is initially at the position 0.280 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.280 m, moving with velocity 0.210 m/s and acceleration 0.310 m/s2. Suppose it moves with constant acceleration for 4.50 s.

(a) Find its position at the end of this time interval.

(b) Find its velocity at the end of this time interval.

Next, assume it moves with simple harmonic motion for 4.50 s and

x = 0

is its equilibrium position. (Assume that the velocity and acceleration is the same as in parts (a) and (b).)

(c) Find its position at the end of this time interval.

(d) Find its velocity at the end of this time interval.

Explanation / Answer

given a particle moves in x direction

xi = 0.28 m

vi = 0.210 m/s

ai = -0.310 m/s/s

ti = 4.5 s

a. xf =xi + vi*ti + 0.5*ai*ti^2

xf = 0.28 + 0.210*4.5 - 0.5*0.310*4.5^2

xf = -1.91375 m

b. vf = vi + ai*ti

vf = 0.210 - 0.310*4.5

vf = -1.185 m/s

c. now,

a = -0.310 m/s/s

for x = 0.28 m

hence

ma = -kx

k/m = -a/x = 0.310/0.28 = 1.1071428

w = sqroot(k/m) = 1.0522 rad/s

hence

the equaiton of motion is

x = Asin(wt + phi)

at t= 0

0.28 = Asin(phi)

0.210 = Awcos(phi)

(0.28)^2/A^2 + 0.210^2/A^2*1.0522^2 = 1

A = 0.34385

hence

0.28 = 0.34385sin(phi)

phi = 54.519 deg

hence

x = 0.34385sin(1.0522t + 54.519 deg)

at t = 4.5 s

x= -0.19322864 m

d. v = 0.299268 m/s

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