5. 5/7 points | Previous Answers My Notes A particle moves along the x axis. It

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5. 5/7 points | Previous Answers My Notes A particle moves along the x axis. It is initially at the position 0.140 m, moving with velocity 0.120 m/s and acceleration -0.270 m/s-. Suppose it moves with constant acceleration for 3.00 s (a) Find the position of the particle after this time 0.715 (b) Find its velocity at the end of this time interval 0.69 We take the same particle and give it the same initial conditions as before. Instead of having a constant acceleration, it oscillates in simple harmonic motion for 3.00 s around the equilibrium position x = 0, Hint: the following problems are very sensitive to rounding, and you should keep all digits in your calculator. (c) Find the angular frequency of the oscillation. Hint: in SHM, a is proportional to x 1.39 IS (d) Find the amplitude of the oscillation. Hint: use conservation of energy 0.165 (e) Find its phase constant 0 if cosine is used for the equation of motion. Hint: when taking the inverse of a trig function, there are always two angles but your calculator will tell you only one and you must decide which of the two angles you need ra (f) Find its position after it oscillates for 3.00 s .157 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m (g) Find its velocity at the end of this 3.00 s time interval 04 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s Submit Answer Save Progress Practice Another Version

Explanation / Answer

here,

Part F:
t = 4.50s

position at time t ,
x = A*Sin(w*t)
x = 0.165*sin(1.39*3)
x = 0.0119 m

Part g:
v = dx/dt
v = 0.165*w*Cos(wt)
v = 0.165*1.39*Cos(1.39*3)
v = 0.2287 m/s or 0.23 m/s( Rounded off)

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