Answer:
D=41.48 ft
[tex]a=54.43\ ft/s^2[/tex]
Explanation:
Given that
y=0.5 x²
Vx= 2 t
We know that
[tex]V_x=\dfrac{dx}{dt}[/tex]
At t= 0 ,x=0
[tex]x=\int V_x.dt[/tex]
At t= 3 s
[tex]x=\int_{0}^{3} 2t.dt[/tex]
[tex]x=[t^2\left\right ]_0^3[/tex]
x= 9 ft
When x= 9 ft then
y= 0.5 x 9² ft
y= 40.5 ft
So distance from origin is
x= 9 ft ,y= 40.5 ft
[tex]D=\sqrt{9^2+40.5^2} \ ft[/tex]
D=41.48 ft
[tex]a_x=\dfrac{dV_x}{dt}[/tex]
Vx= 2 t
[tex]a_x= 2\ ft/s^2[/tex]
At t= 3 s , x= 9 ft
y=0.5 x²
[tex]a_y=\dfrac{d^2y}{dt^2}[/tex]
y=0.5 x²
[tex]\dfrac{dy}{dt}=x\dfrac{dx}{dt}[/tex]
[tex]\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}[/tex]
Given that
[tex]\dfrac{dx}{dt}=2t[/tex]
[tex]\dfrac{dx}{dt}=2\times 3[/tex]
[tex]\dfrac{dx}{dt}=6\ ft/s[/tex]
[tex]a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2[/tex]
[tex]a_y=54\ ft/s^2[/tex]
[tex]a=\sqrt{a_x^2+a_y^2}\ ft/s^2[/tex]
[tex]a=\sqrt{2^2+54^2}\ ft/s^2[/tex]
[tex]a=54.43\ ft/s^2[/tex]
