Respuesta :
Answer:
Part a)
[tex]F = 7.76 N[/tex]
Part b)
[tex]\theta = -137.7 degree[/tex]
Part c)
[tex]\theta = -127.7 degree[/tex]
Explanation:
As we know that acceleration is rate of change in velocity of the object
So here we know that
[tex]x = -19 + t - 3t^3[/tex]
[tex]y = 20 + 7t - 9t^2[/tex]
Part a)
differentiate x and y two times with respect to time to find the acceleration
[tex]a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)[/tex]
[tex]a_x = \frac{d}{dt}(0 +1 - 9t^2)[/tex]
[tex]a_x = -18t[/tex]
[tex]a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)[/tex]
[tex]a_y = \frac{d}{dt}(0 +7 - 18t)[/tex]
[tex]a_y = -18[/tex]
Now the acceleration of the object is given as
[tex]\vec a = (-18t)\hat i + (-18)\hat j[/tex]
at t= 1.1 s we have
[tex]\vec a = -19.8 \hat i - 18 \hat j[/tex]
now the net force of the object is given as
[tex]\vec F = m\vec a[/tex]
[tex]\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)[/tex]
[tex]\vec F = -5.74 \hat i - 5.22 \hat j[/tex]
now magnitude of the force will be
[tex]F = \sqrt{5.74^2 + 5.22^2} = 7.76 N[/tex]
Part b)
Direction of the force is given as
[tex]tan\theta = \frac{F_y}{F_x}[/tex]
[tex]tan\theta = \frac{-5.22}{-5.74} [/tex]
[tex]\theta = -137.7 degree[/tex]
Part c)
For velocity of the particle we have
[tex]v_x = \frac{dx}[dt}[/tex]
[tex]v_x = (0 +1 - 9t^2)[/tex]
[tex]v_y = \frac{dy}{dt}[/tex]
[tex]v_y = (0 +7 - 18t)[/tex]
now at t = 1.1 s
[tex]\vec v = -9.89\hat i - 12.8 \hat j[/tex]
now the direction of the velocity is given as
[tex]\theta = tan^{-1}(\frac{v_y}{v_x})[/tex]
[tex]\theta = tan^{-1}(\frac{-12.8}{-9.89})[/tex]
[tex]\theta = -127.7 degree[/tex]
