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 15 m/s² c 10 m/s² d 5 m/s²

Explanation:

Step1: Recall the formula for acceleration

Acceleration \(a=\frac{\Delta v}{\Delta t}\). 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 velocity \(v = \frac{s}{t}\), where \(s\) is distance and \(t\) is time. For each second, \(v=\frac{5\ m}{1\ s}=5\ m/s\) (constant velocity).

Step3: Calculate the acceleration

If \(v = constant\), then \(\Delta v=v_{final}-v_{initial}=0\). Using the formula \(a = \frac{\Delta v}{\Delta t}\), with \(\Delta v = 0\) (because velocity is constant) and \(\Delta t\) being any non - zero time interval (here \(\Delta t = 1s\) between each second of measurement). So \(a=\frac{0}{1}=0\ m/s^{2}\)

Answer:

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