Step 5 of 6 Find acceleration component in y direction y 05

Step 5 of 6 Find acceleration component in y direction. y 0.5 Differentiate the equation dy dr dt dt Further differentiate the equation dr dr dr substitute, a for d\'y v, for dax a for 51 (2.5x5) a, 37,5ft/s

Solution

He has simply make derivatives of the equation y=0.5*x² with respect to time t.

In the first step he has made differential equation dy/dt = x dx/dt

General rule of differential equation: x*y = x dy/dt + y dx/dt

here 0.5 is constant so, x²(0.5/dt) + (2x*0.5)*(dx/dt) = x dx/dt

here dx/dt will be your velocity in x direction.

further make a step of differential equation for acceleration in y direction

that will be d²y/dt² = (dx/dt)² + x d²x/dt

d²y/dt² shows that it is the second differential of dy/dt ,while second differential of x dx/dt will be like (dx/dt)² + x d²x/dt from the General rule of differential equation: x*y = x dy/dt + y dx/dt

d²y/dt² acceleration in y direction = a_y

dx/dt velocity in x direction = v_x

d²x/dt = acceleration in x direction = a_x

then he has simply put values of a_y,v_x,a_x in last differential equation.

New equation will be like a_y=v_x²+xa_x

and finally put all values and got the answer