A red train traveling at 72 km/hand a green train traveling at $144\mathrm{km}/\mathrm{h}$are headed toward each other along a straight, level track. When they are 950 mapart, each engineer sees the other’s train and applies the brakes. The brakes slow each train at the rate of$1.0\mathrm{m}/{\mathrm{s}}^2$ . Is there a collision? If so, answer yes, give the speed of the red train and the speed of the green train at impact, respectively. If not, answer no and give the separation between the trains when they stop.

The speed of a red train, ${\mathrm{v}}_{0\mathrm{r}}=72\mathrm{km}/\mathrm{h}.$

The speed of a green train,$v_{0\mathrm{g}}=144\mathrm{km}/\mathrm{h}$

The separation between the two trains is 950 m.

The rate at which the speed is decreased,$\mathrm{a}=1.0\mathrm{m}/{\mathrm{s}}^2$

Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion. The kinematic equations relate the velocity, displacement, acceleration, and time to each other.

The final velocity of both trains is zero for the calculation of distance travelled by them before stopping. Using the kinematic equation, For the red train:

$\mathrm{v}={\mathrm{v}}_{0\mathrm{r}}+\mathrm{at}$

(i)

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{v}-{\mathrm{v}}_0}{\mathrm{a}} \\ =\frac{0\mathrm{m}/\mathrm{s}-20\mathrm{m}/\mathrm{s}}{-1\mathrm{m}/{\mathrm{s}}^2} \\ =20\mathrm{s} \end{gathered}$$

The distance traveled by the red train is calculated as,

$$\begin{gathered} ∆\mathrm{x}={\mathrm{v}}_{0\mathrm{r}}\mathrm{t}+\frac{1}{2}\times \left(-1\mathrm{m}/{\mathrm{s}}^2\right){\left(20\mathrm{s}\right)}^2 \\ =200\mathrm{m} \end{gathered}$$

Time taken by green train to come to rest,

$\mathrm{t}=\frac{\mathrm{v}-{\mathrm{v}}_0}{\mathrm{a}}$

Distance travelled by the green train is calculated as,

$$\begin{gathered} ∆\mathrm{x}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ =40\mathrm{m}/\mathrm{s}\times 40\mathrm{s}+\frac{1}{2}\times \left(-1\mathrm{m}/{\mathrm{s}}^2\right){\left(40\mathrm{s}\right)}^2 \\ =800\mathrm{m} \end{gathered}$$

Thus, total distance travelled is $200\mathrm{m}+800\mathrm{m}=1000\mathrm{m},$but given separation is 950 m.

This shows that trains will collide.

The speed of the red train is zero as it travels a distance of 200 m and then stops. The green train moves with speed and travels a distance of 750 m and collides with the red train.

Use the kinematic equation to find the speed of the green train.

$$\begin{gathered} {\mathrm{V}}^2={\left(40\mathrm{m}/\mathrm{s}\right)}^2+2\left(-1\mathrm{m}/{\mathrm{s}}^2\right)\left(750\mathrm{m}\right) \\ =100{\mathrm{m}}^2/{\mathrm{s}}^2 \\ \mathrm{v}=10\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the green train while colliding is, 10m/s.