The velocity of a particle constrained to move along the x-axis as a function of

Question

The velocity of a particle constrained to move along the x-axis as a function of time t is given by:

v(t)=(17/t0)sin(t/t0).

Part A: If the particle is at x=9 m when t=0, what is its position at t = 6t0. You will not need the value of t0 to solve any part of this problem. If it is bothering you, feel free to set t0=1 everywhere.Express your answer to 3 significant figures.

Part B:

Denote instantaneous acceleration of this particle by a(t). Evaluate the expression 9 +v(0)t+a(0)t2/2 at t = 6t0. Note that v(0) and a(0) are the velocity and acceleration at time t=0.

Express your answer to 3 significant figures.

Part C:

Now assume that the particle is at x=0 when t=0, and find its position at t = 0.14t0.

Express your answer to 3 significant figures.

Part D: Evaluate v(0)t+a(0)t2/2 at t =0.14t0 to 3 significant figures. Your answers to parts (c) and (d) should be in good agreement. Try to work out why that is so before submitting the answer to part (d)

Explanation / Answer

a) I guess t_0 is some time constant that I am going to call T
v = (17/T) sin (t/T)
then integrate
x = -17 cos t/T + c
if x = 9 at t = 0 then
9 = -17 + c ======> c = 26
so
x = 26 - 17 cos t/T
at t = 6T
x = 26 - 17 cos6 = 9.09

b) a = dv/dt = 17/T^2 cos t/T
v(0) = 0 since sin 0/T = 0
a(0) = 17/T^2
so
9 + 0 +(8.5/T^2) (36 T^2) = 9 + 306 = 315

c) x = -17 cos t/T + c
if x = 0 at t = 0 then
0 = -17 + c ======> c = 17
so
x = 17 - 17 cos t/T
at t = 0.14T
x = 17 - 17 cos0.14 = 5.07*10^-5

d) a = dv/dt = 17/T^2 cos t/T
v(0) = 0 since sin 0/T = 0
a(0) = 17/T^2
so
0 +(8.5/T^2) (0.0196*T^2) = 0.1666

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