Problem 5 - Applied Mathematics

Question

A particle moves in a straight line and passes through \(\mathrm{O}\), a fixed point on the line with a velocity of \(6 \mathrm{~ms}^{-1}\). The particle moves with a constant retardation of \(2 \mathrm{~ms}^{-2}\) for four seconds and thereafter moves with constant velocity. How long after leaving \(\mathrm{O}\) does the particle return to \(\mathrm{O}\) ? (a) \(3 \mathrm{~s}\) (b) \(8 \mathrm{~s}\) (c) never (d) \(4 \mathrm{~s}\) (e) \(6 \mathrm{~s}\).

Step-by-Step Solution

Answer

8 seconds

Step 1 - Determine the final velocity after retardation

Use the equation of motion: \[ v = u + at \] Here, initial velocity, \( u = 6 \) \( \mathrm{ms}^{-1} \), acceleration, \( a = -2 \) \( \mathrm{ms}^{-2} \) (since it's retardation), and time, \( t = 4 \) seconds. Plug in the values: \[ v = 6 + (-2) \times 4 = 6 - 8 = -2 \] \( \mathrm{ms}^{-1} \). So, the final velocity after retardation is \( -2 \) \( \mathrm{ms}^{-1} \).

Step 2 - Calculate the distance covered during retardation

Use the equation: \[ s = ut + \frac{1}{2}at^2 \] Here, \( u = 6 \) \( \mathrm{ms}^{-1} \), \( a = -2 \) \( \mathrm{ms}^{-2} \), and \( t = 4 \) seconds. Plug in the values: \[ s = 6 \times 4 + \frac{1}{2} \times (-2) \times 4^2 \] \[ s = 24 - 16 = 8 \] meters.

Step 3 - Determine the time to return to point O

Since the final velocity is constant at \( -2 \) \( \mathrm{ms}^{-1} \) after retardation, the distance (8 meters) back to point O will be covered with the same speed. Use the equation: \[ t = \frac{s}{v} \] Here, \( s = 8 \) meters, \( v = 2 \) \( \mathrm{ms}^{-1} \). \[ t = \frac{8}{2} = 4 \] seconds.

Step 4 - Calculate the total time to return to O

The total time is the time for retardation plus the time to return to O. Given retardation time, \( t_1 = 4 \) seconds, and return time, \( t_2 = 4 \) seconds. \[ t_{total} = t_1 + t_2 = 4 + 4 = 8 \] seconds.

Key Concepts

constant retardation

In physics, retardation refers to negative acceleration. This means the velocity of an object decreases over time. If a particle has a constant retardation, it experiences a steady decrease in velocity. In this exercise, the particle starts with an initial velocity of 6 m/s and experiences a constant retardation of 2 m/s². As a result, its velocity decreases uniformly for a duration. It's crucial to remember that retardation has an opposite direction to motion. For this retarding motion, we use equations of motion but replace acceleration with retardation values.

equations of motion

Equations of motion describe how objects move under various forces. There are three main equations of motion in physics. They connect velocity, position, acceleration, and time. For this problem, we've used the equations to find out the final velocity and distance traveled during retardation. The basic form is:

\( v = u + at \) where:

• \( v \) = final velocity
• \( u \) = initial velocity
• \( a \) = acceleration (negative for retardation)
• \( t \) = time

velocity and time calculation

In the exercise, we need to determine how long it takes for the particle to return to its starting point. After experiencing retardation for 4 seconds, the particle's velocity changes to \( -2 \) m/s. The distance it traveled during this period is calculated using: \( s = ut + \frac{1}{2}at^2 \). Once the particle reaches constant velocity, the distance it needs to return to the original point is the same it covered during retardation, which is 8 meters. Using the equation: \( t = \frac{s}{v}\), we find it takes additional 4 seconds to return. Therefore, the total time needed is 4 + 4 = 8 seconds.