Question

A train accelerating uniformly from rest attains a maximum speed of \(40 \text{ m s}^{-1}\) in 20 s. It travels at this speed for 20 s and is brought to rest with uniform retardation in further 40 s. What is the average velocity during the period?

• A. \(80 \text{ m s}^{-1}\)
• B. \(25 \text{ m s}^{-1}\)
• C. \(40 \text{ m s}^{-1}\)
• D. \(30 \text{ m s}^{-1}\)

Hint:

For motion with varying speeds, average velocity is total displacement divided by total time.

Answer 1

The Correct Option is B

Step 1: Distance during acceleration: \[ s_1 = \frac{(0+40)}{2}\times 20 = 400 \text{ m} \]
Step 2: Distance at constant speed: \[ s_2 = 40 \times 20 = 800 \text{ m} \]
Step 3: Distance during retardation: \[ s_3 = \frac{(40+0)}{2}\times 40 = 800 \text{ m} \]
Step 4: Total distance \(= 2000 \text{ m}\) Total time \(= 20+20+40 = 80 \text{ s}\) \[ \text{Average velocity} = \frac{2000}{80} = 25 \text{ m s}^{-1} \]