1) yes, at time of [tex]t=\frac{I}{r_1+r_2}[/tex]
2) only if the car behind is faster; at time [tex]t=\frac{I}{r_1-r_2}[/tex]
Step-by-step explanation:
1)
The separation distance between the two cars is given by
[tex]I-(r_1+r_2)t[/tex]
where
I is the original separation
[tex]r_1[/tex] is the speed of the 1st car
[tex]r_2[/tex] is the speed of the 2nd car
The positive sign between the speeds of the two cars tells us that the two cars are going into opposite directions (so, the relative velocity of the two cars is the sum of the two velocities).
In order for the two cars to meet, this separation distance must go to zero:
[tex]I-(r_1+r_2)t=0[/tex]
Therefore, this will happen at a time t of:
[tex]I=(r_1+r_2)t\\t=\frac{I}{r_1+r_2}[/tex]
2)
If the two cars were going in the same direction, then the relative velocity of the two cars would be (assuming car 1 is behind car 2)
[tex]r_1-r_2[/tex]
So the separation distance between the two cars would be
[tex]I-(r_1-r_2)t[/tex]
Therefore, the two cars would meet at a time t of
[tex]t=\frac{I}{r_1-r_2}[/tex]
However, we must notice that there is a condition for this to occur: in fact, the time can only be positive, so it must be
[tex]r_1>r_2[/tex]
Which means ,the car behind must go faster than the car ahead, otherwise the two cars will never meet.
