Question

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

v(t)=−(18/t0)sin(t/t0).

Part A

If the particle is at x=1 m when t=0, what is its position at t = 7t0. 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 1 +v(0)t+a(0)t2/2 at t = 7t₀. 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 = 0.14t0.

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)

Answer 1

The answer for above problem is explained below.

Given that, victs = -6 h) sinto Integrate on both sides CE) E 7 dt d 2CL) 0 1 + ta t de ace) = simple 7 dt 26 -2 = (-60s (4.] *to)" 2C) -1 = 18 (] -] as ect) = 1 + 18 [cos [ #] - 18 es x(t) = 18 cos (es) -17 At tanto a X (7 to) = 18cas (Ito ) - 17 z x(7 to) = 18 cos (7) - 17

a) x (7to) = 18 (0.7539) - 17 = 13.57 -17 72(7to) = -3.4297 m Ons Acceleration, act) = d V (t) at - o ach are acts = very sin C#] cos (1) - And VCO) = 0 and aco) = - 18/2 Given, Expression is, = 1tvo Ct) + aco) t% 31+0+ ( 3 ) * [7854 = 1 * Uomo

= 1 - 441 = (-440 m I-440m At t= 0.14 to (0.14to) = 18 cos (0.14) -18 = - 0.1761m .: x cooluto) = (-0.176m Day Given, At t = 0.14to Given, At t = 0.1uto = Vot + aco) ¢ +0+1+] (c-toto)_] et Dop10 to 2 -0.176m