Respuesta :
Answer:
(a) [tex]\frac{m}{s^2}[/tex]
(b) [tex]\frac{m}{s^3}[/tex]
(c) 1 s
(d) 20 m
(e) 1 m
(f) [tex]0\frac{m}{s}[/tex]
(g) [tex]-12\frac{m}{s}[/tex]
(h) [tex]-36\frac{m}{s}[/tex]
(i) [tex]-72\frac{m}{s}[/tex]
(j) [tex]-6\frac{m}{s^2}[/tex]
(k) [tex]-18\frac{m}{s^2}[/tex]
(l) [tex]-30\frac{m}{s^2}[/tex]
(m) [tex]-42\frac{m}{s^2}[/tex]
Explanation:
Since x is measured in meters and t in seconds, constants a and b must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean [tex]\frac{m}{s^2}[/tex] for a and [tex]\frac{m}{s^3}[/tex] for b.
We can get the velocity v equation by deriving the position with respect to t, which gives:
[tex]v=6*t-6*t^2[/tex]
And the acceleration a equation by deriving again:
[tex]a=6-12*t[/tex]
Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is v=0, which gives
[tex]6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1[/tex]
For t = 0, x = 0 so the maximun position is archieved at 1 second, which gives x = 1 meter.
For obtaining it's displacement r, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:
[tex]r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m[/tex]
For the remaining questions, we just replace the values of t on the respective equations.
