A particle moves in the first vector of oxy in the plane coordinate system. The trajectory equation is xy = 16, and the change rule of X with time t is x = 4T ^ 2 (t is not equal to 0). The velocity of the particle at t = 1s can be calculated by taking X and Y as M

Vx=dx/dt=8t
y=16/x
Vy=dy/dt=-16*8t/（16t^4）=-8/t^3
When t = 1s, VX = 8, vy = - 8
V = 8 √ 2 m / s, the direction is 45 degrees below the slope

The motion equation of the particle in the XY plane is: x = 2T, y = T2. Calculate the acceleration in the first second, the tangential acceleration, the normal phase acceleration and the orbit radius

x=2t,y=t2,
x'=2 y'=2t
x''=0 y''=2
Total acceleration a = 2m / S
Velocity v = √ (2 ^ 2 + (2t) ^ 2) = 2 √ (1 + T ^ 2)
When t = 1, V1 = 2 √ 2m / S
Tangential acceleration V '= 2 √ (1 + T ^ 2))' = 2T / √ (1 + T ^ 2)
When t = 1, the tangential acceleration AT1 = √ 2m / S ^ 2
Normal acceleration an1 = √ (a ^ 2-at1 ^ 2) = √ 2m / S ^ 2
Track radius r = V1 ^ 2 / an1 = 4 √ 2m / S ^ 2m / S

The motion equation of the particle on the XY plane is x = 3T + 5, y = 1 / 2 * (T ^ 2) + 3t-4, and the displacement from 1s to 2S is obtained

Substituting t = 1 into the coordinate equation, we can get the particle position a: x = 8, y = - 0.5, and substituting t = 2 into B, x = 11, y = 4
AB vector (3,4.5) is the displacement from a to B!

A particle moves in the first vector of oxy in the plane coordinate system. The trajectory equation is xy = 16, and the change rule of X with time t is x = 4T ^ 2 (t is not equal to 0). The velocity of the particle at t = 1s can be calculated by taking X and Y as M