Problem 20 - Master Resource Book in JEE Main Physics

Question

The motion of a body falling from rest in a resisting medium is described by the equation $$ \frac{d v}{d t}=A-B v $$ where \(A\) and \(B\) are constants. Then (a) initial acceleration of the body is \(A\) (b) the velocity at which acceleration becomes zero is \(A / B\) (c) the velocity at any time \(t\) is \(\frac{A}{B}\left(1-e^{B t}\right)\) (d) All of the above are wrong

Step-by-Step Solution

Answer

All given options are correct; therefore, (d) is wrong.

Step 1: Understand the Initial Condition

We are given the differential equation \( \frac{d v}{d t}=A-B v \), which describes the rate of change of velocity \( v \) with time \( t \). The body falls from rest, implying the initial velocity \( v(0) = 0 \).

Step 2: Calculate the Initial Acceleration

The initial acceleration can be found by substituting \( v = 0 \) into the differential equation: \( \frac{d v}{d t} = A - B \times 0 = A \). Thus, the initial acceleration of the body is \( A \).

Step 3: Determine Velocity at Zero Acceleration

The acceleration becomes zero when \( \frac{d v}{d t} = 0 \), implying \( A - Bv = 0 \). Solving for \( v \), we get \( v = \frac{A}{B} \). This is the velocity at which acceleration becomes zero.

Step 4: Solve the Differential Equation for Velocity at Time t

Rearrange the differential equation to separate variables: \( \frac{d v}{A - Bv} = dt \). Integrating both sides, we get \( -\frac{1}{B} \ln |A-Bv| = t+C \). Solving for \( v \) and using the initial condition \( v(0) = 0 \), we obtain \( v(t) = \frac{A}{B} (1 - e^{-Bt}) \).

Step 5: Ascertain the Correct Option

Option (a) states the initial acceleration is \( A \), which matches our findings in Step 2. Option (b) states the velocity at zero acceleration is \( \frac{A}{B} \), which we verified in Step 3. Option (c) provides the velocity at time \( t \), which we derived in Step 4. Hence, all these statements are correct.

Key Concepts

Motion in a Resisting Medium

When a body moves through a resisting medium, such as air or water, the resistance from the medium impacts its motion. This resistance often acts as a damping force opposing the motion of the object. In our exercise, the equation \( \frac{d v}{d t}=A-B v \) defines this phenomenon for a body falling from rest.

• Resisting Medium: The medium exerts a force that depends on the velocity of the falling object. The resisting force is proportional to the velocity, which is captured by the term \(-Bv\) in our equation.
• Differential Equation: The presence of a differential equation indicates that a function and its derivatives are involved, representing physical quantities like velocity and time.

This kind of equation helps us predict how the body's speed changes over time, making it a key element in describing motion within physics.

Initial Conditions in Differential Equations

Initial conditions are crucial in solving differential equations as they provide the specific information required to find unique solutions. In the context of motion in a resisting medium, we start with the object at rest, which translates to an initial velocity of zero.

• Importance of Initial Conditions: Initial conditions, like \(v(0) = 0\), allow us to solve the equation appropriately for this particular scenario.
• Calculating Initial Values: From the differential equation \(\frac{d v}{d t}=A-B v\), substituting \(v=0\) provides immediate acceleration as \(A\).

Knowing the initial velocity helps us plot the motion of the object over time, ensuring we have a complete and accurate picture of the system's dynamics.

Velocity-Time Relationship

The relationship between velocity and time is pivotal in understanding how quickly or slowly an object moves through a resisting medium. For the given differential equation, the velocity as a function of time was derived by integrating the equation.

• Equation Transformation: By separating variables and integrating, you obtain \( v(t) = \frac{A}{B} (1 - e^{-Bt}) \).
• Exponential Decay: The term \(e^{-Bt}\) represents an exponential decay, illustrating how the velocity changes over time. This decay is characteristic of motion resistance, where the speed approaches a terminal velocity of \(\frac{A}{B}\).

This relationship provides insights into how the object accelerates initially but gradually reaches a stable velocity as the resisting medium's force balances other forces exerted on the object.