Respuesta :
Answer:
a) [tex] z'(t) =v(t) = -13t[/tex]
Now we can replace the velocity for t=1.75 s
[tex] v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}[/tex]
For t = 3.0 s we have:
[tex] v(3.0s) = -13*3.0 =-39 \frac{m}{s}[/tex]
b) [tex] v_{avg}= \frac{z_f - z_i}{t_f -t_i}[/tex]
And we can find the positions for the two times required like this:
[tex] z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m[/tex]
[tex] z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m[/tex]
And now we can replace and we got:
[tex] V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}[/tex]
Explanation:
The particle position is given by:
[tex] z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0[/tex]
Part a
In order to find the velocity we need to take the first derivate for the position function like this:
[tex] z'(t) =v(t) = -13t[/tex]
Now we can replace the velocity for t=1.75 s
[tex] v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}[/tex]
For t = 3.0 s we have:
[tex] v(3.0s) = -13*3.0 =-39 \frac{m}{s}[/tex]
Part b
For this case we can find the average velocity with the following formula:
[tex] v_{avg}= \frac{z_f - z_i}{t_f -t_i}[/tex]
And we can find the positions for the two times required like this:
[tex] z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m[/tex]
[tex] z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m[/tex]
And now we can replace and we got:
[tex] V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}[/tex]
