A particle moves along the x axis. It is initially at the position 0.290 m, movi

Question

A particle moves along the x axis. It is initially at the position 0.290 m, moving with velocity 0.210 m/s and acceleration -0.440 m/s2. Suppose it moves with constant acceleration for 5.10 s. (a) Find the position of the particle after this time. 4.36 (b) Find its velocity at the end of this time interval. -2.034 m/s Next, assume it moves with simple harmonic motion for 5.10 s and x0 is its equilibrium position. (Assume that the velocity and acceleration is the same as in parts (a) and (b).) (c) Find its position. 0.336 Your response differs from the correct answer by more than 10%. Double check your calculations, m (d) Find its velocity at the end of this time interval. 0.44 Your response differs from the correct answer by more than 100%, m/s

Explanation / Answer

Assume the position for SHM is described by
x = A cos(t - )

v = dx/dt = -A sin(t - )
a = dv/dt = -A² cos(t - )

Given the initial conditions, solve for , , and A.
a/x = -²
(-0.440 m/s²)/(0.290 m) = -²
= 1.123 rad/s

a/v = cot(t - )
(-0.440 m/s²)/(0.210 m/s) = (1.123 rad/s)cot(-)
tan = 0.587 (tan and cot are odd functions... no negative signs)
= 0.53075463 rad

x = A cos(t - )
0.290 m = A cos(-)
0.290 m = A cos(0.531 rad)
A = 0.336 m

(c)
x = (0.336 m) cos[(1.123 rads^-1)(5.10 s) - 0.531]
x = 0.156 m

(d)
v = -(0.336 m)(1.123 rads^-1) sin[(1.123 rads^-1)(5.10 s) - 0.531]
v = 0.336 m/s

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