Answer to Question #111864 in Physics for gaurav

Question

The acceleration of Point A is defined by the relation a = -1.8sinkt , where a and t are expressed in m/s2 and seconds, respectively, and k = 3 rad/s. Knowing that x = 0 and v = 0.6 m/s when t = 0, determine the velocity and position of Point A when t = 0.5 s.

Answer 1

Given:

$a=-1.8\sin(3t){\:\rm m/s^2},\\ x(0)=0{\:\rm m},\\ v(0)=0.6{\:\rm m/s},\\ t=0.5\:\rm s.$

The velocity of a point

$$v(t)=\int a(t)dt=\\ =-1.8\int \sin(3t)dt=0.6\cos(3t)+C$$ $$v(0)=0.6\cos(0)+C=0.6+C=0.6,\; C=0.$$

Hence,

$$v(t)=0.6\cos(3t),\\ v(0.5)=0.6\cos(3\times 0.5)=0.04\:\rm m/s.$$

The position of a point

$$x(t)=\int v(t)dt=\\ =0.6\int \cos(3t)dt=0.2\sin(3t)+C$$ $$x(0)=0.2\sin(0)+C=C=0,\; C=0.$$

Hence,

$$x(t)=0.2\sin(3t),\\ x(0.5)=0.2\sin(3\times 0.5)=0.2\:\rm m.$$