Question

A particle moves along the x-axis so that its velocity at any time t/geq0 is given by v(t) = 1 - sin(2t). (a) Determine the expression for acceleration at any time t. (b) Find all values t, 0 <t<2, for which the particle is at rest. (c) Determine the expression for the position Jf the particle at any timet if x(0) = 0.

Answer 1

Given

$V(t) = 1 - sin(2\pi t)$

a) Accerelation function of time.

$acceleration = \frac{\mathrm{d} V(t)}{\mathrm{d} x} = - 2\pi cos(2\pi t)$

$acceleration = - 2\pi cos(2\pi t)$

b) time at particle is at rest. i.e means V(t) = 0

V(t) = 0 = 1 sin (2nt

$sin(2\pi t) = 1$

$2\pi t= sin^{-1}(1) = \pi/2 , 3\pi/2$

$t = \frac{1}{4} , \frac{3}{4}$

c) Position of particle function of time.x(t)

$V(t) = \frac{\mathrm{d} x(t)}{\mathrm{d} t}$

integrate above equation we get x(t)

$x(t) = \int V(t)dt$

$x(t) = \int (1 - sin(2\pi t))dt$

$x(t) = t + \frac{cos(2\pi t)}{2\pi} + c$.........c is the constant of integration.

Given x(0) = 0

put in equation x(t) t = 0

$then C = -\frac{1}{2\pi}$

Position of particle = $x(t) = t + \frac{cos(2\pi t)}{2\pi} - \frac{1}{2\pi}{\color{Red} }$