Question

The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along the x axis is vx=(2t)ft/s, where t is in seconds, determine the particle's distance from the origin O. t = 3 s. When t=0, x=0, y=0. Determine the magnitude of its acceleration when t = 3 s.

Answer 1

Answer:

D=41.48 ft

$a=54.43\ ft/s^2$

Explanation:

Given that

y=0.5 x²                      

Vx= 2 t

We know that

$V_x=\dfrac{dx}{dt}$

At t= 0 ,x=0  

$x=\int V_x.dt$

At t= 3 s

$x=\int_{0}^{3} 2t.dt$

$x=[t^2\left\right ]_0^3$

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

$D=\sqrt{9^2+40.5^2} \ ft$

D=41.48 ft

$a_x=\dfrac{dV_x}{dt}$

Vx= 2 t

$a_x= 2\ ft/s^2$

At t= 3 s , x= 9 ft

y=0.5 x²    

$a_y=\dfrac{d^2y}{dt^2}$

y=0.5 x²    

$\dfrac{dy}{dt}=x\dfrac{dx}{dt}$

$\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}$

Given that

$\dfrac{dx}{dt}=2t$

$\dfrac{dx}{dt}=2\times 3$

$\dfrac{dx}{dt}=6\ ft/s$

$a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2$

$a_y=54\ ft/s^2$

$a=\sqrt{a_x^2+a_y^2}\ ft/s^2$

$a=\sqrt{2^2+54^2}\ ft/s^2$

$a=54.43\ ft/s^2$