Answer:
[tex]X - Xo = 54m[/tex]
k = 1/18
Explanation:
Data:
a = -k[tex]t^{2}[/tex][tex]\frac{m}{s^{2} }[/tex]
to = 0s Vo = 12m/s
t = 6s the particle chage it's moviment, so v = 0 m/s
We know that acceleration is the derivative of velocity related to time:
[tex]a = \frac{dV}{dT}[/tex]
rearranging...
[tex]a*dT = dV[/tex]
Then, we must integrate both sides:
[tex]\int\limits^f_i {dV} \, dV =-k \int\limits^f_i {t^{2} } \, dT[/tex]
[tex]V - Vo = -k\frac{t^{3} }{3}[/tex]
V = 0 because the exercise says that the car change it's direction:
[tex]0 - 12 = -k\frac{6^{3} }{3}[/tex]
k = 1/6
In order to find X - Xo we must integer v*dT = dX
[tex]V - Vo = -k\frac{t^{3} }{3}[/tex]
so...
[tex](Vo -k\frac{t^{3} }{3})dT = dX[/tex]
[tex]\int\limits^f_i {dX} \, dX = \int\limits^f_i {Vo -k\frac{t^{3} }{3} } \, dT[/tex]
integrating...
[tex]X - Xo = Vot -k\frac{t^{4} }{12}[/tex]
[tex]X - Xo = 12*6 -\frac{1}{6}* \frac{6^{4} }{12}[/tex]
X - Xo = 54m
