Answer:
It will take the first car 9.52 seconds since its starting
Explanation:
Please see image to understand the position of the cars for the race, and the tangential velocity vector and radius associated with each of them.
Notice that for these cars to pass each other they have to be at the same angle in their respective circumferences. Therefore we need to create equations that describe the angle subtended by each in their movement. Since we assume they move at constant speed, we need to convert their tangential velocities into angular velocities (the equation is simple if the angular velocity is in radians/sec. Therefore we will also stick to radians for the angle covered in the motion.
Relationship between radius, tangential velocity and angular velocity (in rad/sec) is given by: [tex]v_t=w*r\\w=\frac{v_t}{r}[/tex]
where [tex]v_t[/tex] is the tangential velocity, [tex]w[/tex] is the angular velocity in radians/sec), and [tex]r[/tex] is the radius.
Therefore for the first car (blue in the picture), its angular velocity will be: [tex]w_1=\frac{25}{20} \frac{rad}{s} =1.25\frac{rad}{s}[/tex]
For the second car (red in the picture), its angular velocity will be: [tex]w_2=\frac{60}{30} \frac{rad}{s} =2\frac{rad}{s}[/tex]
Now the angle described by each moving at constant angular speed will be given by:
[tex]\Theta(t)=\Theta_i+w*t\\[/tex]
where [tex]\Theta_i[/tex] is the starting angle (if not zero). For the first car, according to our picture that has the red car at the origin (angle zero), this initial angle is 180 degrees ([tex]\pi[/tex] in radians). So the angular equation will be:
[tex]\Theta_1(t)=\pi +1.25\frac{rad}{s} *t[/tex]
For the second car, the starting angle will be zero, but remember also that this car starts 2 seconds after the previous one, so we include that 2 second difference in the time (t) which counts when the blue car starts the race:
[tex]\Theta_2(t)=0+2\frac{rad}{s} *(t-2)[/tex]
The crossing takes place when the two angles are the same, that is when [tex]\Theta_1(t)=\Theta_2(t)[/tex], so we set that equation and solve for the time (t):
[tex]\pi +1.25\frac{rad}{s}*t =2\frac{rad}{s} *(t-2s)\\\\\pi +1.25*t=2*t-4\\\pi+4=2*t-1.25*t\\\pi +4=0.75*t\\t=\frac{\pi +4}{0.75} = 9.5221s[/tex]
This is the time (counted from the starting of car 1) that it takes for the cars to pass each other.
