Question

20 a train travels 5 meters in the 1st second of travel, 5 meters again during the 2nd second of travel, and 5 meters again during the third second. what is its acceleration? a 0 m/s² b 5 m/s² c 15 m/s² d 10 m/s²

Explanation:

Step1: Recall the formula for acceleration

Acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_{final}-v_{initial}\). Also, if the distance traveled in each successive second is the same, the velocity is constant.

Step2: Analyze the velocity

Since the train travels 5 meters in the \(1^{st}\) second, 5 meters in the \(2^{nd}\) second, and 5 meters in the \(3^{rd}\) second. The displacement per second is constant. Using the formula \(v = \frac{s}{t}\) (where \(s\) is displacement and \(t\) is time), for each second \(v=\frac{5\ m}{1\ s}=5\ m/s\). The initial velocity \(v_{i}\) and final velocity \(v_{f}\) (after each second - interval) are equal. So, \(\Delta v=v_{f} - v_{i}=0\).

Step3: Calculate acceleration

Using \(a=\frac{\Delta v}{\Delta t}\), with \(\Delta v = 0\) and \(\Delta t=1\ s\), we get \(a=\frac{0}{1}=0\ m/s^{2}\)

Answer:

A. \(0\ m/s^{2}\)