Question

A particle is thrown in upward direction with initial velocity of $60 \mathrm{~m} / \mathrm{s}$. Find average speed and average velocity after 10 seconds. \(\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\) (A) \(26 \mathrm{~ms}^{-1}, 16 \mathrm{~ms}^{-1}\) (B) \(26 \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\) (C) \(20 \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\) (D) \(15 \mathrm{~ms}^{-1}, 25 \mathrm{~ms}^{-1}\)

Short answer

The average speed and average velocity after 10 seconds are \(20 \mathrm{~ms}^{-1}\) and \(10 \mathrm{~ms}^{-1}\), respectively. So, the correct option is (C).

Step-by-step solution

Calculate the final position after 10 seconds

To find the final position (s) after 10 seconds, we need to use the following equation of motion: \(s = ut + \frac{1}{2}at^2\) where u is the initial velocity (60 m/s), t is the time (10 seconds), and a is the acceleration (-10 m/s^2, negative because gravity acts downward) Putting the values, we get: \(s = (60)(10) - \frac{1}{2}(10)(10)^2\) \(s = 600 - 500\) \(s = 100 m\)

Calculate the final velocity after 10 seconds

To find the final velocity (v) after 10 seconds, we'll use the following equation of motion: \(v = u + at\) where u is the initial velocity (60 m/s), t is the time (10 seconds), and a is the acceleration (-10 m/s^2) Putting the values, we get: \(v = 60 - (10)(10)\) \(v = 60 - 100\) \(v = -40 m/s\)

Calculate the total distance

To calculate the total distance traveled by the particle, we need to consider both the upward and downward paths. The downward path will have the same distance as the upward path. Therefore, the total distance will be: \(Total Distance = 2 \times s = 2 \times 100\) \(Total Distance = 200 m\)

Calculate the average speed

The average speed is the total distance traveled divided by the total time. In this case, the total distance is 200 m and the total time is 10 seconds: \(Average Speed = \frac{Total Distance}{Total Time} = \frac{200}{10}\) \(Average Speed = 20 m/s\)

Calculate the average velocity

The average velocity is the change in displacement divided by the total time. The displacement after 10 seconds is 100 m (final position) with upward direction: \(Average Velocity = \frac{Displacement}{Total Time} = \frac{100}{10}\) \(Average Velocity = 10 m/s\) This implies that the correct option is (C) \(20 \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\).