A particle moves along the x axis. It is initially at the position 0.160 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.160 m, moving with velocity 0.240 m/s and acceleration -0.380 m/s2. Suppose it moves with constant acceleration for 3.80 s

(a) Find the position of the particle after this time.

=    m

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

=      m/s

Next, assume it moves with simple harmonic motion for 3.80 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.

=   m

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

=    m/s

Explanation / Answer

x = x0+(u*t) + (1/2)*a*t^2...

x = 0.16+(0.24*3.8) + (1/2)*0.38*3.8^2...

x = 3.8156 m....
B) v = u+at = 0.24+(0.38*3.8) = 1.684 ms^-1....

w = 2*pi/T = 6.284/3.8 = 1.66...

x = A*cos(wt) = 0.16*cos(1.66*3.8) = 0.159 m...

v = A*w*sin(wt) = 0.16*1.66*sin(1.66*3.8) = 0.0065 ms^-1

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