Question

What is the time required by a train of length of \(350 \mathrm{~m}\) to cross an electric pole with a speed of \(70 \mathrm{~m} / \mathrm{s}\) ? (a) \(3 \mathrm{~s}\) (b) \(3.5 \mathrm{~s}\) (c) \(5 \mathrm{~s}\) (d) \(10 \mathrm{~s}\)

Short answer

Answer: 5 seconds

Step-by-step solution

Write down the given information

The length of the train is \(350 \mathrm{~m}\), and its speed is \(70 \mathrm{~m} / \mathrm{s}\).

Write the distance formula

Distance \(=\) Speed \(\times\) Time or \(d=s \times t\).

Insert the known values into the formula

In our case, \(d=350\) and \(s=70\); therefore, the equation becomes \(350=70 \times t\).

Solve for the time

In order to find the time, we need to divide both sides of the equation by 70: $$ t=\frac{350}{70}. $$

Simplify and find the answer

When we divide 350 by 70, we get: $$ t=\frac{350}{70}=5. $$ Therefore, the time required for the train to cross the electric pole is \(5 \mathrm{~s}\), which corresponds to option (c).

Key concepts

Motion along Straight Line

When we talk about motion along a straight line, we are looking at how an object travels in one dimension - a straight path without alterations in direction. This is different from circular motion or motion in two or three dimensions. In the given exercise, the train is moving in a straight line as it crosses the electric pole.

In these types of problems, there's a specific focus on the object's speed, direction, and distance traveled. The speed of the object is constant when it moves with uniform motion in a straight line. This means that the object covers equal distances in equal intervals of time. This is precisely what happens with the train moving at a steady speed of 70 m/s.

The length of the train, in this context, is the distance it travels to completely cross a reference point, like the electric pole. By understanding these elements, you can predict how long it would take for the object to traverse another point or find out how far it can travel in a certain time frame.

Problem Solving Techniques

Solving time-speed-distance problems efficiently requires a structured approach, and this includes using logical steps to find the solution. First, we identify all the given information, like the train's speed and the distance it needs to cover - which in this scenario is its own length, 350 m.

The key formula to use here is the distance formula:

• Distance (d) = Speed (s) × Time (t).

By rearranging this formula, you can solve for the unknown variable, which in our case is time (t).

After filling in the known values, we change the formula to accommodate our needs. This requires isolating the unknown variable by manipulating the equation appropriately, such as dividing both sides by the train's speed to get the time. The division itself is straightforward and reduces the problem down to a simple calculation.

Ensuring careful execution of each step and confidence in basic arithmetic can enhance your ability to solve similar problems more effortlessly.

Basic Algebra

Basic algebra is crucial when solving time, speed, and distance problems. It provides the tools to manipulate equations and find unknown values, which is essential in real-world physics applications like predicting how long it takes for a train to pass by a given point.

In our problem, you use algebra to solve the equation for time (t):

• Start by knowing the equation setup, which involves expressing distance as a product of speed and time.
• Substitute the given values - here, distance as 350 m and speed as 70 m/s - into this formula.
• To isolate the unknown (time), divide both sides by the known variable, which is the speed.
• Finally, simplify by performing the arithmetic operation, resulting in the time it takes for the train to pass a pole.

Understanding these steps reinforces fundamental algebra skills, helping students not only in similar mathematical problems but also in developing analytical reasoning for broader academic success.