Answer:
140.04 m from the stop.
Explanation:
We can divide the motion in two parts:
First, we have an uniformly accelerated motion, using the following formula we calculate the distance traveled in this part:
[tex]v_f^2=v_0^2+2ax_1\\x_1=\frac{v_f^2}{2a}\\x_1=\frac{(32\frac{m}{s})^2}{2(4\frac{m}{s^2})}\\x_1=128m[/tex]
In the second part, the car decelerates, but we don't know the value of this deceleration. We have to calculate the time spend in the first part, in order to know the time spend on second part:
[tex]v_f=v_0+at_1\\t_1=\frac{v_f}{a}=\frac{32\frac{m}{s}}{4\frac{m}{s^2}}\\t_1=8s[/tex]
Recall that total time is 15 s
[tex]t=t_1+t_2\\t_2=t-t_1\\t_2=15s-8s=7s[/tex]
Now, we calculate the deceleration:
[tex]a=\frac{-v_0}{t_2}\\a=\frac{-32\frac{m}{s}}{7s}\\a=-4.57\frac{m}{s^2}[/tex]
Finally, we find the distance traveled in the second part. In this way, we can know how far does the car travel.
[tex]x_2=\frac{-v_0^2}{2a}\\x_2=\frac{-(32\frac{m}{s})^2}{2(-4.57\frac{m}{s^2})}\\x_2=112.04m\\x=x_1+x_2\\x=128m+112.04m\\x=140.04m[/tex]
