NPCIL Trainee Acceleration — practice questions

1. The stem describes a velocity that is changing over time, which is exactly what acceleration measures. 2. The introduction to this topic states: "Pushing on the gas pedal results in acceleration because the velocity of the car increases, and acceleration is defined as a change in velocity." 3. Displacement describes a change in position, not velocity; average speed and reference frame do not describe a rate of change at all. 4. So the correct term for this change in velocity is acceleration. _Source: OpenStax Physics Ch 3 "Acceleration", p.97 §3.1 Acceleration_

1. Average acceleration = change in velocity / change in time. 2. The train starts from rest, so the change in velocity is simply the final velocity: 8.33 m/s − 0 m/s = 8.33 m/s. 3. Divide by the elapsed time: a = 8.33 m/s / 20.0 s ≈ 0.417 m/s². 4. 8.33 m/s² wrongly treats the velocity value itself as the acceleration 20.0 m/s² wrongly treats the time value as the acceleration 1.50 m/s² does not correspond to either given number divided correctly. 5. So the train's average acceleration is about 0.417 m/s², directed the same way as its motion since it is speeding up. _Source: OpenStax Physics Ch 3 "Acceleration", p.101 Worked Example §3.1 Acceleration_

1. Average acceleration = (final velocity − initial velocity) / elapsed time. 2. Here the final velocity is 0 and the initial velocity is 8.333 m/s, so the change in velocity is 0 − 8.333 = −8.333 m/s. 3. Divide by the elapsed time: a = −8.333 m/s / 8.00 s ≈ −1.04 m/s². 4. The negative sign correctly shows the acceleration points opposite to the train's direction of travel, consistent with it slowing down (a deceleration). 5. The +1.04 m/s² option gets the magnitude right but the wrong sign; the ±0.417 m/s² options mistakenly reuse the earlier problem's 20.0 s time instead of this problem's 8.00 s. _Source: OpenStax Physics Ch 3 "Acceleration", p.101 Worked Example §3.1 Acceleration_