Question

PLEASE HELP

Answer 1

Answer:

a) t =12[s]; b) x = 348[m]

Explanation:

We can solve this problem using the following kinematics equations:

a)

$v_{f}= v_{o}+a*t$

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:

$t=\frac{v_{f} -v_{o} }{a} \\t = \frac{12-6}{0.5} \\t=12[s]$

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].

$v_{f}^{2}= v_{o}^{2}+2*a*x_{1} \\therefore\\x_{1} =\frac{v_{f}^{2}-v_{o}^{2}}{2*a} \\x_{1} =\frac{12^{2}-6^{2}}{2*.5}\\x_{1} =108[m]$

The we can calculate the second displacement for the constant velocity:

$x_{2} =x_{o}+v*t_{2} \\ x_{2} =0+12*(20)\\x_{2} =240[m]$

x = x1 + x2

x = 108 + 240

x = 348[m]