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)=(19/t0)sin(t/t0).

Part A

If the particle is at x=6 m when t=0, what is its position at t = 9t0. 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.

Part B

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

Part C

Now assume that the particle is at x=0 when t=0, and find its position at t = 0.13t0.

Part D

Evaluate v(0)t+a(0)t2/2 at t =0.13t0 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) We know that
distance = Velocity*time = (-19/to)Sin(t/to) dt
Integrating this with in the limit 0 to 9to
= (19)(Cos(t/to) - Cos(0) ) = 19 (Cos 9 - Cos1)
we get distance = 0.231 m
If the particle is initially at 6 m then final position = 0.231 = 6.231 m
(b) We know that
S = Si +ut +0.5at2
S = 6 + 0 + 0.5a(o)t2
a = dV/dt = -(19/to)Cos(t/to)(1/to)
a = -19/to2   at t = 0
S = 6 + 0.5(-19/to2)t2
(c) similarly like 1
distance
= 19 (Cos0.13 - cos1 ) = 0.0028 m
(d) a(o) = 19/to2
   = 0 + 19/to2*(1/2)*(0.13to)2 = 0.161 m/s2

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