A particle is traveling along the parabolic path y 025 x2 I

A particle is traveling along the parabolic path y = 0.25. x^2. If x = 8 m, upsilon_x = 8 m/s, and a_x = 4 m/s^2 when t = 2 s, determine the magnitude of the particle\'s velocity and acceleration at this instant.

Solution

>> Equation of Path : y = 0.25*x²

>> Differentiating above equation with respect to time,

=> dy/dt = 0.25*2*x*dx/dt

As, Vy = dy/dt and Vx = dx/dt

=> Above Equation becomes:

Vy = 0.5*x*Vx

>> Now, at t = 2 sec, x = 8 m, Vx = 8 m/s

Putting all values above,

=> Vy = 0.5*8*8 = 32 m/s

=> Vy = 8 m/s

=> Net Velocity, V = [ Vx² + Vy² ]^1/2 =  [ 8² + 32² ]^1/2 = 33 m/s ........REQUIRED VELOCITY .....ANSWER....

>> As, Above Equation is:

Vy = 0.5*x*Vx

>> Now, Differentiating above with respect to tine:

=> Ay = 0.5*Vx² + 0.5*x*Ax [ where, Ax = dVx/dt and Ay = dVy/dt ]

>> Now, at t = 2 sec, x = 8 m , Vx = 8 m/s and Ax = 4 m/s2

Putting all values above,

=> Ay = 0.5*8*8 + 0.5*8*4

=> Ay = 48 m/s²

>> So, Net Acceleration, A = [ Ax² + Ay² ]^1/2 =  [ 4² + 48² ]^1/2 = 48.2 m/s² ........REQUIRED ACCELERATION .....ANSWER....