In a particular Cartesian coordinate system, a particle has coordinates (33%) Pr

Question

In a particular Cartesian coordinate system, a particle has coordinates

(33%) Problem 2: In a particular Cartesian coordinate system, a particle has coordinates x(t) 2sin(3t) + C, where t is in seconds. x is in meters, and C is a constant to be determined by the data. At t = 0 the particle was at x = 1 m. 14% Part (a) Find the value of constant C, in meters cosO sin cotanasin0 acosO acotansinh0 coshO anh cotanhO END Degrees Radians CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 2 Feedback: 0%-deduction per feedback. - 14% Part (b) Find the instantaneous velocity, in meters per second, at t-0.95 s. 14% Part (c) Find the instantaneous velocity, in meters per second, at t 1.9 s. 14% Part (d) Find the instantaneous velocity, in meters per second, at t = 2.7 s. - 14% Part (e) Find the instantaneous acceleration, in meters per square second, at t 0.95 s 14% Part (f) Find the instantaneous acceleration, in meters per square second, at t = 1.9 s. - 14% Part (g) Find the instantaneous acceleration, in meters per square second, at t-2.7 s

Explanation / Answer

part a:

at t=0, particle was at x=1.

==>2*sin(0)+C=1

==>C=1

part b:

instantaneous velocity=dx/dt=6*cos(3*t)

at t=0.95 s, instantaneous velocity=6*cos(3*0.95)=-5.7467 m/s

part c:

at t=1.9 seconds, instantaneous velocity=6*cos(3*1.9)=5.0083 m/s

part d:

at t=2.7 seconds, instantaneous velocity=6*cos(3*2.7)=-1.4613 m/s

part e:

instantaneous acceleration=d^2x/dt^2

=-18*sin(3*t)

at t=0.95 seconds, instantaneous acceleration=18*sin(3*0.95)

=-5.1746 m/s^2

part f:

instantaneous acceleration=d^2x/dt^2

=-18*sin(3*t)

at t=1.9 seconds, instantaneous acceleration=18*sin(3*1.9)=9.9123 m/s^2

part g:

instantaneous acceleration=d^2x/dt^2

=-18*sin(3*t)

at t=2.7 seconds, instantaneous acceleration=18*sin(3*2.7)=-17.458 m/s^2

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