Q62P

Question

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track (Fig. P2.62). The freight train is traveling at 15.0 m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of $0.100 m / s^{2}$ in a direction opposite to the train’s velocity, while the freight train continues with constant speed. Take x = 0 at the location of the front of the passenger train when the engineer applies the brakes. (a) Will the cows nearby witness a collision? (b) If so, where will it take place? (c) On a single graph, sketch the positions of the front of the passenger train and the back of the freight train.

Figure P2.62

Step-by-Step Solution

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Answer

Answer is not given in the drive.

1Step 1: Identification of the given data

The given data can be expressed below as:

The engineer of the passenger train is travelling at 25.0 m/s.
The caboose of the passenger train is 200 m ahead.
The speed of the freight train is 15.0 m/s.
The constant acceleration of the engineer of the passenger train is $0.100 m / s^{2}$ .
The location of the front of passenger train is x = 0.

2Step 2: Significance of the Newton’s law of motion for the passenger train

This law elucidates that an object will continue to be at rest or at uniform motion unless an external force acts upon the object.

The equation of the motion gives the time and the distance of the collision along with a graph.

3Step 3: Determination of the time and the distance of the collision with the graph

a) As soon as the brake has been applied, the time t and the displacement x of the train both becomes 0. The initial velocity of the train $v_{i}$ becomes 25.0 m/s and the acceleration of the train a becomes $- 0.100 m / s^{2}$ .

Let, t be the time taken for slowing down the train to 15 m/s, hence, from the Newton’s first law, the equation of the velocity of the train can be expressed as:

$v_{f} = v_{i} + a t$

Substituting the values in the above expression, we get-

$\begin{aligned} 15 m / s = 25.0 m / s - {(- 0.100 m / s^{2})}^{2} t \\ t = 100 s \end{aligned}$

Hence, the distance travelled by the passenger train during t = 100 s is:

$d_{1} = v_{t} t + \frac{1}{2} a t^{2}$

Substituting the values in the above equation, we get-

$\begin{aligned} d_{1} = (25.0 m / s) \times 100 s + \frac{1}{2} \times (- 0.100 m / s^{2}) \times (100 s)^{2} \\ = 2000 m \end{aligned}$

On the other hand, the velocity of the freight train is v = 15 m/s, and the train takes 100 s to slow down. Hence, the distance travelled by the train is expressed as:

$\begin{aligned} d_{2} = v t \\ = 15 m / s \times 100 s \\ = 1500 m \end{aligned}$

As the caboose of the freight train is 200 m ahead, so the distance travelled by the train is 1500 m + 200 m = 1700 m.

Thus, the collision will happen as the passenger train will hit the freight train before slowing down to 15 m/s.

b) Analysing from the above figure, the collision is going to happen when $d_{1} = d_{2} + 200 m$ … i)

As we know that $d_{1} = v_{t} t + \frac{1}{2} a t^{2}$ ,

And vt,

So, from the equation i), we get-

$\begin{aligned} 25.0 m / s \times t + 0.5 \times (- 0.100 m / s^{2}) \times t^{2} = 15 m / s \times t + 200 m \\ - 0.05 t^{2} + 10 t - 200 m = 0 \end{aligned}$