Answer:
Explanation:
Given the position of a particle in space at a time t expressed as;
[tex]r(t) = \frac{\sqrt{3} }{3}t i +(t-16t^{2} )j \\[/tex]
Velocity of the body at t = 0 will be expressed as shown;
[tex]v(t) = dr/dt = \frac{\sqrt{3} }{3}i + (1-32t)j\\ v(0) = \frac{\sqrt{3} }{3}i + j\\\\acceleration\ at \ t=0;\\\\a(t) = dv(t)/dt = -32j\\\\a(0) = -32j[/tex]
To get the angle between the velocity and acceleration at t = 0, we will use the formula for calculating the dot product of two vectors.
[tex]v.a = |v||a|cos\theta\\[/tex]
[tex]v.a = (\sqrt{3}/3i + j) . (-32j)\\v.a = -32\\\\|v| = \sqrt{(\sqrt{3}/3)^2 + 1^2} \\|v| = \sqrt{1/3+1}\\ |v| = \sqrt{4/3}\\ |v| = 2/\sqrt{3}\\ \\|a| = \sqrt{(-32)^2}\\ \\|a| = 32[/tex]
Substituting the values given into the formula, we will have;
[tex]-32 = 32*2/\sqrt{3} cos\theta\\\\ -1 = \frac{2}{\sqrt{3} } } cos\theta\\\\-\sqrt{3} = 2cos\theta\\\\cos\theta = -\sqrt{3}/2\\\\ \theta = cos^{-1} \frac{-\sqrt{3} }{2}\\ \\\theta = -30^0\\\\[/tex]
Since cos is negative in the second quadrant, the angle between the velocity and acceleration will be 180 - 30 = 150°
