Answer:
The entering car is going to catch up with the other car after 20.76 seconds.
Step-by-step explanation:
The car leaving the pit area is moving with a constant accelaration, then we can calculate the speed it has when entering the main speedway using the following equation:
[tex]s_1=s_0+a*t_1[/tex]
Where [tex]s_0[/tex] is the initial speed (which is zero given that the car starts froms rest), [tex]a[/tex] is the car's accelaration and [tex]t_1[/tex] is the time passed until it reaches the main speedway.
[tex]s_1=s_0+a*t_1=a*t_1[/tex]
[tex]s_1=a*t_1=(5.1 \frac{m}{s^2})(3.6s)=18.36 \frac{m}{s}[/tex]
For an object travelling at a constant accelation, the displacement [tex]x[/tex] at a time [tex]t[/tex] can be calculate using the following equation:
[tex]x=x_1+(s_1*t)+\frac{1}{2}at^2[/tex] (equation 1)
Let's consider [tex]x_1[/tex] the point where the first car enters the main speedway. If [tex]x_2[/tex] is the point where this car catch up with other one after [tex]t_2[/tex] seconds, we may rewrite the equation 1 like this:
[tex]x_2=x_1+(s_1*t_2)+\frac{1}{2}a(t_2)^2[/tex]
[tex]x_2=(s_1*t_2)+\frac{1}{2}a(t_2)^2[/tex] (equation 2)
In other hand, the second car is travelling at a constant speed [tex](v)[/tex] when it meets the entering car. Then, its displacement [tex](d_2)[/tex] after [tex]t_2[/tex] seconds can be calculated using the following formula:
[tex]d_2=v*t_2[/tex] (equation 3)
The entering car is going to catch up with the other one when [tex]x_2=d_2[/tex], so when can find how much time this will require equaling equations 2 and 3 and isolating [tex]t_2[/tex]
[tex]v*t_2=(s_1*t_2)+\frac{1}{2}a(t_2)^2[/tex]
[tex]0=(s_1*t_2)+\frac{1}{2}a(t_2)^2-v*t_2[/tex]
[tex][\frac{1}{2}(a*t_2)-(v-s_1)]*t_2=0[/tex]
[tex]\left \{ {{t_2=0} \atop {\frac{1}{2}(a*t_2)-(v-s_1)=0}} \right.[/tex]
[tex]\frac{1}{2}(a*t_2)-(v-s_1)=0[/tex]
[tex]t_2=\frac{2}{a}(v-s_1)=\frac{2}{5.1 \frac{m}{s^2}}(71.3\frac{m}{s}-18.36\frac{m}{s})=20.76 s[/tex]
So, the entering car is going to catch up with the other car after 20.76 seconds.
