Answer:
a) [tex]v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}[/tex]
b) [tex]v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}[/tex]
c) [tex]a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }[/tex]
Explanation:
The knowable variables are
[tex]d_{t}=25m[/tex]
[tex]t_{1}=1.3 s[/tex]
[tex]t_{2}=3.9 s[/tex]
[tex]t_{3}=5.5 s[/tex]
Since the three traffic signs are equally spaced, the distance between each sign is [tex]\frac{25}{3} m [/tex]
a) [tex]v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}[/tex]
b) [tex]v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}[/tex]
Since we know the velocity in two points and the time the car takes to pass the traffic signs
c) [tex]a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }[/tex]
