Answer:
[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]
Explanation:
We are given that
[tex]y=(0.05x^2)[/tex]
Where x (in m)
x component of velocity, [tex]v_x=-4m/s[/tex]
x- component of acceleration,[tex]a_x=-1.2 m/s^2[/tex]
x=3 m
We have to find the y-component of velocity and acceleration at this instant.
Differentiate w.r.t. time
[tex]\frac{dy}{dt}=0.05\times 2x\frac{dx}{dt}[/tex]
[tex]\frac{dy}{dt}=0.1x\frac{dx}{dt}=0.1xv_x[/tex]
By using the formula
[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]
[tex]v_x=\frac{dx}{dt}[/tex]
Substitute the values
[tex]v_y=\frac{dy}{dt}=0.1(3)\times (-4)=-1.2m/s[/tex]
Again differentiate w.r.t. t
[tex]a_y=\frac{dv_y}{dt}=0.1v_x+0.1x\frac{d(v_x)}{dt}[/tex]
Substitute the values
[tex]a_y=0.1(-4)+0.1(3)(-1.2)=-0.4-0.36=-0.76m/s^2[/tex]
Where [tex]a_x=\frac{dv_x}{dt}[/tex]
Hence, the y- component of velocity and acceleration
[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]
