Kinematics' equations

We can say that the distance between B and C is:

$Vcar2 \times t = 120$

$Vcar1 \times (t+1) = 120$

( $t$ refers to the time it took for car2 to get from B to C )

We need to experss $Vcar2$ in proportion to $Vcar1$ to solve these two equations.

The time it took for car3 to drive 120 km from B to C + 40 km back Is equal to The time it took for car1 to drive 80 km from B in the direction of C :

$\frac { 80 }{ Vcar1 } =\frac { 120+40 }{ Vcar3 }$

$80\times Vcar3=160\times Vcar1$

$Vcar3 = 2 \times Vcar1$

We can see that $Vcar3$ is twice as big as $Vcar1$ .

Since we know that the time intervals of the cars leaving from A to B are equal, and that the cars reach B simultaneously we can write:

$AB = Vcar1 \times T = Vcar3 \times (T-0.5T) = Vcar2 \times (T-xT)$

$(T$ refers to the time it took for car1 to get from A to B )

It took car3 only half the time it took car1 to get from A to B because it had twice the speed.

$(T-xT)$ which is the time interval of car2 (faster than car1 ) has to be 2 times smaller than the interval of car3 $(T-0.5T)$ in order for the time intervals between the 3 cars to be equal (equal to $0.25T$ ).

(Equal interavals between the cars can be: $0.25T$ in that case $Vcar3 > Vcar2$ , or $0.5T$ which makes : $(T-xT)=T-1T=0$ in other words - not possible ).

$(T-xT) = \frac { (T-0.5T) }{ 2 }$

$T-xT=-0.25T$

$x=0.25$

From substituting $x$ we can get the connection we wanted between $Vcar1$ and $Vcar2$ :

$Vcar1 \times T = Vcar2 \times (T-0.25T)$

$Vcar1=0.75\times Vcar2$

$Vcar2=\frac { Vcar1 }{ 0.75 }$

Then we go back to our first two equations and replace $Vcar2$ :

$\frac { Vcar1 }{ 0.75 } \times t = 120$

$Vcar1 \times (t+1) = 120$

And now we solve for $Vcar1$ :

$t = \frac { 90 }{ Vcar1 }$

$Vcar1 \times ( \frac { 90 }{ Vcar1 } + 1) = 120$

$90 + Vcar1 = 120$

$Vcar1 = 30$ km/hr