Problem 23 - Master Resource Book in JEE Main Physics

Question

A particle starts from rest and travels a distance \(s\) with uniform acceleration, then it travels a distance \(2 s\) with uniform speed, finally it travels a distance \(3 s\) with uniform retardation and comes to rest. If the complete motion of the particle in a straight line then the ratio of its average velocity to maximum velocity in (a) \(6 / 7\) (b) \(4 / 5\) (c) \(3 / 5\) (d) \(2 / 5\)

Step-by-Step Solution

Answer

The ratio of average velocity to maximum velocity is \(\frac{6}{7}\), which is option (a).

Step 1: Analyze the Motion

The motion consists of three phases: acceleration, uniform speed, and deceleration to rest. The particle starts from rest, accelerates while traveling a distance \(s\), continues at a constant speed for a distance \(2s\), and decelerates for a distance \(3s\).

Step 2: Maximum Velocity Calculation

Let the maximum velocity achieved by the particle be \(v_m\). This maximum velocity is reached at the end of the first phase (i.e., after traveling distance \(s\)). The particle then travels at this maximum velocity \(v_m\) over the next segment of distance \(2s\).

Step 3: Average Velocity Calculation

The average velocity \(v_{avg}\) for the entire motion is defined as the total displacement divided by the total time taken. Total displacement = \(6s\). Since each segment of the journey is either at acceleration, constant speed, or constant deceleration, the main task is to derive the time taken for each part.

Step 4: Calculate Time for Each Segment

Using kinematic equations:1. **First Segment**: - Distance = \(s\), Starting velocity \(u = 0\), Final velocity \(v_m\). - Using \(v^2 = u^2 + 2as\) gives \(v_m^2 = 0 + 2as\) implies \(a = \frac{v_m^2}{2s}\). - Time for this segment \(t_1 = \frac{v_m}{a} = \frac{2s}{v_m}\).2. **Second Segment**: - Constant velocity \(v = v_m\), Distance = \(2s\). - Time for this segment \(t_2 = \frac{2s}{v_m}\).3. **Third Segment**: - Distance = \(3s\), Final velocity \(v = 0\). - Using \(v^2 = u^2 + 2as\) and rearranging to \(t_3 = \frac{2(3s)}{v_m}\) when stopping.

Step 5: Calculate Total Time and Velocity Ratios

Total time \(T = t_1 + t_2 + t_3 = \frac{2s}{v_m} + \frac{2s}{v_m} + \frac{3s}{v_m} = \frac{7s}{v_m}\).The average velocity is \(v_{avg} = \frac{6s}{T} = \frac{6s}{(7s/v_m)} = \frac{6v_m}{7}\). Therefore, the ratio \(\frac{v_{avg}}{v_m} = \frac{6}{7}\).

Step 6: Validate the Calculation

Reexamine the calculations for each step, ensuring all algebraic manipulations support the solution. This confirms that the proportion of average velocity to the maximum velocity is consistent with our finding.

Key Concepts

Uniform Acceleration

In kinematics, uniform acceleration refers to a constant change in velocity over a period of time. Picture a car speeding up steadily as it leaves a stoplight—it doesn't jerk forward but gains speed smoothly. This means at every second, the same amount of velocity is added. The particle from our exercise experiences this while covering its first segment of distance, denoted as \(s\), starting from rest (zero initial velocity).
To determine the acceleration \(a\), we use the kinematic equation for motion:

• \(v^2 = u^2 + 2as\)

Here, \(v\) is the final velocity, \(u\) is the initial velocity, which is zero, \(a\) is acceleration, and \(s\) is the distance covered. Given these parameters, it's calculated that \(v_m^2 = 2as\), leading to \(a = \frac{v_m^2}{2s}\).
Time taken for this uniform acceleration can be found using the formula:

• \(t_1 = \frac{2s}{v_m}\)

This time accounts for how long the particle takes to reach its maximum speed, contributing to understanding the overall motion and subsequent segments.

Average Velocity

Average velocity is a crucial concept in kinematics as it represents the overall change in position over time. Imagine setting out on a long journey—your average speed or velocity describes how far you traveled on average per hour, combining all stops and bursts of speed into one value. In our exercise, the particle's average velocity \(v_\text{avg}\) applies to the entire trajectory through varied motion.
Formulated through:

• \(v_\text{avg} = \frac{\text{total displacement}}{\text{total time}}\)

For the particle that completed a 3-phase motion totaling a displacement of \(6s\), this interplay of distance and the collectively tallied time (\(T\)) simplifies into:

• \(v_\text{avg} = \frac{6s}{(7s/v_m)} = \frac{6v_m}{7}\)

The end result is a simplified view of its motion, especially in regards to pace and time expenditure—illustrating the importance of average velocity as a measure of efficiency and speed comparison.

Maximum Velocity

Maximum velocity \(v_m\) is the highest speed achieved by the particle during its journey. You can think of it as peaking on a speedometer where the needle can't go any higher at that moment. Once the particle accelerates and reaches the final speed of the first segment, it maintains this pace before entering a deceleration phase.
In the problem, this is the velocity used for traveling a distance of \(2s\) at a uniform speed—a phase where speed doesn't change. To find this maximum velocity, we rely on the equations used in the acceleration phase, as it's the endpoint of the initial acceleration:

• \(v_m\) is discussed and not computed directly in the provided steps, but it's vital for deriving other aspects like time and average velocity.

The comprehension of maximum velocity is essential as it represents the potential top speed in a body's motion, impacting how the average velocity is calculated.