Question

A particle is projected vertically upwards with velocity \(30 \mathrm{~ms}^{-1}\). Find the ratio of average speed and instantaneous velocity after 6 s. \(\left[\mathrm{g}=10 \mathrm{~ms}^{-1}\right]\) (A) \((1 / 2)\) (B) 2 (C) 3 (D) 4

Short answer

Based on the step-by-step solution, the short answer is as follows: The ratio of average speed and instantaneous velocity after 6 seconds is \(0/(-30) = 0\). However, there might be an error in the problem's statement or the answer choices, as 0 is not among the given options.

Step-by-step solution

Write down the given values and equations

Given Initial velocity, u = 30 m/s Acceleration, a = -10 m/s^2 (since acceleration due to gravity is downward) Time, t = 6 s We need to find the ratio: (Average speed)/(Instantaneous velocity)

Calculate the final velocity

To find the final velocity after t=6s, we can use the following equation: v = u + a*t Substituting the values, we get: v = 30 + (-10)*6 v = 30 - 60 = -30 m/s (The negative sign indicates that the particle is moving downward after 6 seconds)

Calculate the average speed

The average speed is defined as the total distance covered divided by the total time taken. Since the particle is projected upwards and then falls back to its highest point, from where it starts to fall back, we should take into account only the upward distance traveled during the 6 seconds interval. We can use the following equation for distance (s): s = u*t + 0.5*a*t^2 Substituting the values, we get: s = 30*6 - 0.5*10*(6^2) s = 180 - 180 = 0 Since the distance is 0, the average speed is also 0.

Calculate the instantaneous velocity

We have already found the final velocity in step 2, which is the instantaneous velocity after 6 seconds. So, v_inst = -30 m/s

Find the ratio of the average speed and the instantaneous velocity

Now, we can calculate the ratio between the average speed and the instantaneous velocity: (Average speed)/(Instantaneous velocity) = 0/(-30) = 0 But the given options do not have 0 as an option, so there might be an error in the problem's statement or the answer choices.

Key concepts

Projectile Motion

Projectile motion is a form of motion where an object is launched into the air and moves under the influence of gravity. In the exercise given, a particle is projected vertically upwards. While it's common to think of projectiles like cannonballs or arrows that follow a parabolic path, this example simplifies things a bit by focusing on vertical motion. Here’s what to keep in mind:

• In vertically projected motion, the initial velocity is directed upward, counteracting gravity initially.
• The gravitational force continually accelerates the particle downwards at a rate of 10 m/s² (as given in the problem).
• Because the particle is moving upward initially, its velocity will decrease until it reaches a maximum height. At this point, its velocity becomes zero before it starts descending.

In the example, we only deal with the first 6 seconds, which implies the particle might have already reached its peak and begun falling back down. This understanding helps in computing velocities accurately, as seen in the steps provided.

Instantaneous Velocity

Instantaneous velocity refers to the velocity of an object at a particular point in time. This is different from average velocity, which considers the whole duration of motion.

• In calculus terms, instantaneous velocity is the derivative of the position function with respect to time. But here we're using physics equations.
• In the exercise, the instantaneous velocity at time t = 6 seconds was calculated using the straightforward kinematic equation: \[ v = u + at \].
• Thus, the particle's velocity turns out to be \( -30 \text{ m/s} \), which means it's moving downward at that moment.

This negative sign is crucial. It informs us of the direction of motion. Instantaneous velocity doesn't care about where you started, only about how quickly you're moving, and in which direction.

Average Speed

Average speed is an important concept in motion. It is calculated by dividing the total distance traveled by the time it took to travel that distance.

• In this scenario, average speed doesn’t take direction into account, only the magnitude of motion.
• The peculiar aspect of the solution to this exercise is that the particle traveled up and then back down, covering the same vertical distance, which mathematically resulted in zero net displacement after the 6-second period.
• This zero displacement implies the average speed also calculates as zero since there's no distance effectively "covered" over the time period.

This irony in calculations underscores the importance of distinguishing between displacement (a vector) and total path length, which affects the average speed calculation. The key takeaway is that despite average speed including all motion, the peculiar nature of vertical projection can result in seemingly paradoxical results if not carefully interpreted.