Question

The relation between velocity and position of a particle is \(\mathrm{V}=\mathrm{Ax}+\mathrm{B}\) where \(\mathrm{A}\) and \(\mathrm{B}\) are constants. Acceleration of the particle is \(10 \mathrm{~ms}^{-2}\) when its velocity is \(\mathrm{V}\), How much is the acceleration when its velocity is \(2 \mathrm{~V}\) (A) \(20 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-1}\) (C) \(5 \mathrm{~ms}^{-2}\) (D) 0

Short answer

The acceleration when the velocity is 2V is \(20 \, \mathrm{m/s}^2\), which is option (A).

Step-by-step solution

Find the acceleration formula from the velocity formula

Differentiate the given velocity formula V = Ax + B with respect to time to find the acceleration formula. Recall that acceleration is the time derivative of velocity, so we are looking for a'(t), where a(t) represents acceleration, and V(t) represents velocity. a(t) = \(\frac{dV(t)}{dt} = \frac{d(Ax + B)}{dt}\)

Differentiate the velocity function

Differentiate the given velocity function with respect to time: \(\frac{dV}{dt} = \frac{d(Ax)}{dt} + \frac{d(B)}{dt}\) Since A and B are constants, the derivative of B with respect to time is 0. Also, to find the derivative of Ax with respect to time, we can use the chain rule: \(\frac{d(Ax)}{dt} = A\frac{dx}{dt}\) The derivative of x with respect to time is the velocity, so \(\frac{dx}{dt} = V\). Therefore, the acceleration function is: a(t) = A * V

Find the acceleration at V

We are given that the acceleration is 10 m/s² when the velocity is V. Use this information to determine the value of A in the acceleration function: 10 m/s² = A * V To find A, divide both sides by V: A = \(\frac{10 \, \mathrm{m/s}^2}{V}\)

Calculate the acceleration when the velocity is 2V

Now that we've found the value of A in terms of V, we'll calculate the acceleration when the velocity is 2V. Simply replace V in the acceleration function with 2V: a'(t) = A * (2V) a'(t) = \(\frac{10 \, \mathrm{m/s}^2}{V}\) * (2V) The V's cancel out, leaving: a'(t) = 20 m/s²

Determine the answer from the given choices:

The acceleration when the velocity is 2V is 20 m/s², which corresponds to option (A). Hence, the correct answer is (A) \(20 \, \mathrm{m/s}^2\).

Key concepts

Differentiation

Differentiation is a powerful concept in calculus that helps us understand how things change. It involves calculating the derivative, which is a measure of how a function's output changes as its input changes. In physics, differentiation is essential as it allows us to find rates of change, like velocity or acceleration.
In the context of the problem, we start with a velocity equation \( V = Ax + B \). By differentiating this equation with respect to time, we can find the particle's acceleration, which is the rate of change of velocity.

• The derivative of a constant, like \( B \), is zero because constants don't change.
• Using the chain rule helps differentiate more complex functions like \( Ax \), where \( A \) is constant and \( x \) is variable.

So, differentiation provides the tools needed to reveal how fast or slow changes are happening in various scenarios.

Velocity

Velocity is central to kinematics as it describes an object’s speed in a specific direction. It differs from speed because speed is scalar (magnitude only), while velocity is a vector, including both magnitude and direction.
In the given exercise, velocity \( V \) is defined by the formula \( V = Ax + B \), which relates it to the position \( x \) of a particle. As the position changes over time, the velocity may also change accordingly. When analyzing motion, understanding velocity helps describe how fast an object moves.

• Positive velocity indicates movement in one direction, while negative indicates the opposite direction.
• If velocity changes, it indicates acceleration is present.

Thus, velocity is crucial for understanding the dynamics of movement and how different factors interact.

Chain Rule

The chain rule is a technique in calculus used to find the derivative of composite functions. It tells us how to differentiate functions nested inside one another, like \( f(g(x)) \).
In our exercise, the chain rule helps find how \( Ax \), a component of the velocity function, changes with time. Since \( A \) is a constant, we focus on \( x \), the position.

• We see \( \frac{d(Ax)}{dt} = A \frac{dx}{dt} \).
• The \( \frac{dx}{dt} \) represents velocity \( V \).

By utilizing the chain rule, we find \( a(t) = A \cdot V \), linking acceleration to velocity through differentiation. This technique simplifies understanding how parts of functions relate in changing conditions.

Kinematics

Kinematics is the study of motion without considering the forces that cause it. It focuses on quantities like position, velocity, and acceleration.
The exercise explores kinematics by relating velocity \( V = Ax + B \) and its impact on acceleration. We seek to determine how changes in velocity, influenced by position \( x \), result in certain accelerations.

• Acceleration tells us how quickly an object's velocity changes.
• By understanding initial conditions, we predict motion behavior over time.

Kinematics allows us to model real-world situations by knowing initial parameters, such as velocity and acceleration, leading to predictions about motion outcomes.