There is a moving particle in the oxy plane, and the de motion function is r = 3Ti + 10 ^ 2J (SI) With an explanation

There is a moving particle in the oxy plane, and the de motion function is r = 3Ti + 10 ^ 2J (SI) With an explanation

V=dr/dt=d(3ti+10^2 j)/dt=3i
The particle moves in a straight line with uniform velocity, and the tangential and normal accelerations are 0

When the acceleration of the particle in the plane is known to be at the origin, the acceleration of the particle in the plane is known to be at any time,

v=at=（5t^2i+3j）t=5t^3i+3tj
s=1/2at²=1/2（5t^2i+3j）t²=5/2t^4i+3/2t²j
The position is (5 / 2T ^ 4,3 / 2T & # 178;)

The motion equation of a particle in the oxy plane is x = 6T, y = 4T ^ 2-8 (SI). Then when t = 1s, what are the tangential acceleration and normal acceleration of the particle?

V1 = 6 in X direction, V2 = 8t in Y direction, VV = v1v1 + v2v2 = 36 + 64tt tangential acceleration, A1 = DV / dt = 1 / 2 * 64 * 2 * t / (36 + 64t * t) note that this bracket represents the root sign temporarily, the root sign cannot be typed = 6.4t = 6.4 total acceleration, dv2 / dt = a = 8 normal acceleration, A2 * A2 = aa-a1 * A1 = 8 * 8-6.4 * 6.4 = 4.8 * 4.8 normal acceleration