Respuesta :
Answer:
a. position function of the coin:
[tex]s=-16t^2+1344[/tex]
Now the velocity function:
[tex]v=-32t[/tex]
b. [tex]v_{avg}=-112\ m.s^{-1}[/tex]
c. [tex]v_3=-96\ m.s^{-1}[/tex] & [tex]v_4=-128\ m.s^{-1}[/tex]
d. [tex]t=9.165\ s[/tex]
e. [tex]v_f=293.285\ m.s^{-1}[/tex]
Explanation:
Given:
- height of dropping the silver dollar, [tex]h=1344\ ft[/tex]
Given position function associated with free falling objects:
[tex]s=-16t^2+v_0t+s_0[/tex]
here:
[tex]s_0=[/tex] initial height
[tex]v_0=[/tex]initial velocity
[tex]t=[/tex] time of observation
a)
position function of the coin:
[tex]s=-16t^2+1344[/tex]
∵the object is dropped it was initially at rest
Now the velocity function:
[tex]v=\frac{d}{dt} s[/tex]
[tex]v=-32t[/tex]
b)
we know average velocity is given as:
[tex]\rm v_{avg}=\frac{total\ displacement}{total\ time}[/tex]
Displacement in the given interval:
[tex]s_{_{3-4}}=s_4-s_3[/tex]
[tex]s_{_{3-4}}=(-16\times 4^2+1344)-(-16\times 3^2+1344)[/tex]
[tex]s_{_{3-4}}=-112\ ft[/tex]
Now,
[tex]v_{avg}=\frac{-112}{4-3}[/tex]
[tex]v_{avg}=-112\ m.s^{-1}[/tex]
c)
Instantaneous velocity at t = 3 s:
[tex]v_3=-32\times 3[/tex]
[tex]v_3=-96\ m.s^{-1}[/tex]
Instantaneous velocity at t = 4 s:
[tex]v_4=-32\times 4[/tex]
[tex]v_4=-128\ m.s^{-1}[/tex]
d)
At ground we have s=0:
Put this in position function:
[tex]0=-16t^2+1344[/tex]
[tex]t=9.165\ s[/tex]
e)
Velocity of the coin at impact:
[tex]v_f=-32\times 9.165[/tex]
[tex]v_f=293.285\ m.s^{-1}[/tex]
