The equation of motion of a point moving in the xoy plane is x = 2T, y = 19-2t ^ 2. When is the vector of the particle perpendicular to the velocity?

The equation of motion of a point moving in the xoy plane is x = 2T, y = 19-2t ^ 2. When is the vector of the particle perpendicular to the velocity?

The slope of the vector of a particle k = Y / X
The equation of motion is y = 19-1 / 2 * x ^ 2
The slope of velocity y '= - 2x
Then: k = - 1 / Y‘
y/x=1/(2x)
y=1/2
The equation of motion leads to X

If a particle moves in the oxy plane and the kinematic equations x = 2T and y = 19-2t ^ 2, the average velocity of the point is given in the second second second

X is 4 in the second second, y is the position of Y in the second second minus the initial position, which is 19-11 = 8
The displacement is 4 ^ 2 + 8 ^ 2, root = 4 times root 5
The average velocity is s / T = 2 times root 5

If the motion equation of a particle is s = t - (2t-1) 2, the instantaneous velocity at t = 1s is ()
A. -1B. -3C. 7D. 13

When t = 1, the instantaneous velocity v = s ′| t = 1 = 5-8T | t = 1 = - 3