Problem 15 - Master Resource Book in JEE Main Physics

Question

A particle starts from the origin and moves along the \(X\)-axis such that the velocity at any instant is given by \(4 t^{3}-2 t\), where \(t\) is in second and velocity is in \(\mathrm{ms}^{-1} .\) What is the acceleration of the particle when it is \(2 \mathrm{~m}\) from the origin? (a) \(10 \mathrm{~ms}^{-2}\) (b) \(12 \mathrm{~ms}^{-2}\) (c) \(22 \mathrm{~ms}^{-2}\) (d) \(28 \mathrm{~ms}^{-2}\)

Step-by-Step Solution

Answer

The acceleration is \(22 \mathrm{~ms}^{-2}\).

Step 1: Write the Velocity Function

The velocity of the particle is given by the function: \[ v(t) = 4t^3 - 2t \]

Step 2: Determine the Relationship Between Displacement and Time

The displacement from the origin as a function of time, \(s(t)\), can be found by integrating the velocity function. Using the fundamental theorem of calculus: \[ s(t) = \int v(t) \, dt = \int (4t^3 - 2t) \, dt \]

Step 3: Integrate the Velocity Function to Find Displacement

Perform the integration: \[ s(t) = \int (4t^3 - 2t) \, dt = \frac{4}{4}t^4 - \frac{2}{2}t^2 + C = t^4 - t^2 + C \]Since the particle starts from the origin, \(s(0) = 0\), this implies \(C = 0\). Thus,\[ s(t) = t^4 - t^2 \]

Step 4: Solve for Time when Displacement is 2m

We set the displacement equation to 2 to find \(t\): \[ t^4 - t^2 = 2 \]This simplifies to:\[ t^4 - t^2 - 2 = 0 \]This is a quadratic in \(t^2\). By substitution \(x = t^2\), we solve: \[ x^2 - x - 2 = 0 \]Which factors to: \[ (x-2)(x+1) = 0 \]This gives \(x = 2\) or \(x = -1\). Hence, \(t^2 = 2\) (since time is not negative).\[ t = \sqrt{2}\]

Step 5: Write the Acceleration Function

Acceleration \(a(t)\) is the derivative of velocity \(v(t)\): \[ a(t) = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2 \]

Step 6: Find the Acceleration at the Required Displacement

Substitute \(t = \sqrt{2}\) into the acceleration function:\[ a(\sqrt{2}) = 12(\sqrt{2})^2 - 2 = 12 \times 2 - 2 = 24 - 2 = 22 \, \mathrm{ms}^{-2} \]

Key Concepts

Velocity Function and Its Role in Motion

Understanding how velocity functions allow us to determine how fast something is moving is key to kinematics. In our exercise, the velocity of a particle along the X-axis is described by a specific function: \[ v(t) = 4t^3 - 2t \]This function tells us how the velocity changes over time:

• The term \(4t^3\) suggests the velocity increases more rapidly as time goes on, given the cubic nature of \(t^3\).
• The \(-2t\) term shows an opposing effect, slowing the increase in velocity.

Much like a speedometer reads speed at each moment, this velocity function provides the speed at any time \(t\). Knowing this helps in understanding how the particle moves moment by moment.
By examining and analyzing the velocity function, we can gain a deeper understanding of the motion characteristics of the particle.

Calculating Acceleration from Velocity

Acceleration is essentially how quickly the velocity changes with respect to time. It's the rate of change of velocity. In calculus terms, acceleration is the derivative of velocity. For our given problem, the velocity function is:\[ v(t) = 4t^3 - 2t \]To find the acceleration function, we differentiate the velocity function:\[ a(t) = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2 \]This tells us that acceleration is a function of time, \(t\). The formula specifically indicates:

• Acceleration is influenced significantly by the term \(12t^2\), doubling its effect over time.
• The \(-2\) indicates a slight reduction in acceleration.

With this, we can conclude that the acceleration is not constant; it grows with time. For instance, when \(t = \sqrt{2}\), the acceleration becomes:\[ a(\sqrt{2}) = 12 \times 2 - 2 = 22 \, \mathrm{ms}^{-2} \]Thus, understanding how to calculate acceleration allows us to predict how motion will change in the future.

Displacement-Time Relationship: Integrating Velocity

Displacement refers to how far an object has moved from its starting position over time. Unlike distance, it considers the direction of motion. To find displacement as a function of time from the velocity function, we integrate the velocity function:\[ s(t) = \int v(t) \, dt = \int (4t^3 - 2t) \, dt \]This calculation gives:\[ s(t) = t^4 - t^2 + C \]Where \(C\) is the constant of integration. Given our conditions (the particle starts at the origin):

• \(s(0) = 0\) tells us \(C = 0\).
• Thus, \(s(t) = t^4 - t^2\).

Using this displacement function, we can solve for the specific time \(t\) when the particle is 2 meters from the origin:\[ t^4 - t^2 = 2 \]By solving, we find \(t = \sqrt{2}\).
Grasping how to determine displacement from velocity is crucial, as it allows us to understand where the particle is at any given time along its path.