Question

The equation Bold r (t )equals(StartFraction StartRoot 3 EndRoot Over 3 EndFraction t )Bold i plus (t minus 16 t squared )Bold j is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t equals 0.

Answer 1

150°

Step-by-step explanation:

Given the position of a particle in space at a time t expressed as;

`r(t) = (√(3) )/(3)t i +(t-16t^(2) )j \\`

Velocity of the body at t = 0 will be expressed as shown;

`v(t) = dr/dt = (√(3) )/(3)i + (1-32t)j\\ v(0) = (√(3) )/(3)i + j\\\\acceleration\ at \ t=0;\\\\a(t) = dv(t)/dt = -32j\\\\a(0) = -32j`

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.

`v.a = |v||a|cos\theta\\`

`v.a = (√(3)/3i + j) . (-32j)\\v.a = -32\\\\|v| = \sqrt{(√(3)/3)^2 + 1^2} \\|v| = √(1/3+1)\\ |v| = √(4/3)\\ |v| = 2/√(3)\\ \\|a| = √((-32)^2)\\ \\|a| = 32`

Substituting the values given into the formula, we will have;

`-32 = 32*2/√(3) cos\theta\\\\ -1 = (2)/(√(3) ) } cos\theta\\\\-√(3) = 2cos\theta\\\\cos\theta = -√(3)/2\\\\ \theta = cos^(-1) (-√(3) )/(2)\\ \\\theta = -30^0\\\\`

Since cos is negative in the second quadrant, the angle between the velocity and acceleration will be 180 - 30 = 150°