Answer:
a) t =12[s]; b) x = 348[m]
Explanation:
We can solve this problem using the following kinematics equations:
a)
[tex]v_{f}= v_{o}+a*t[/tex]
where:
vf = final velocity = 12 [m/s]
vo= initial velocity = 6 [m/s]
a = acceleration = 0.5[m/s^2]
t = time [s]
Now clearing the time t, we have:
[tex]t=\frac{v_{f} -v_{o} }{a} \\t = \frac{12-6}{0.5} \\t=12[s][/tex]
b)
We can calculate the displacement for the first 12 [s] then using the equation for the constant velocity we can calculate the other displacement for the 20[s].
[tex]v_{f}^{2}= v_{o}^{2}+2*a*x_{1} \\therefore\\x_{1} =\frac{v_{f}^{2}-v_{o}^{2}}{2*a} \\x_{1} =\frac{12^{2}-6^{2}}{2*.5}\\x_{1} =108[m][/tex]
The we can calculate the second displacement for the constant velocity:
[tex]x_{2} =x_{o}+v*t_{2} \\ x_{2} =0+12*(20)\\x_{2} =240[m][/tex]
x = x1 + x2
x = 108 + 240
x = 348[m]
