Question

A particle is moving in a straight line with initial velocity of \(200 \mathrm{~ms}^{-1}\) acceleration of the particle is given by \(\mathrm{a}=3 \mathrm{t}^{2}-2 \mathrm{t}\). Find velocity of the particle at 10 second. (A) \(1100 \mathrm{~ms}^{-1}\) (B) \(300 \mathrm{~ms}^{-1}\) (C) \(900 \mathrm{~ms}^{-1}\) (D) \(100 \mathrm{~ms}^{-1}\)

Short answer

The velocity of the particle at 10 seconds is (A) \(1100\mathrm{~ms}^{-1}\).

Step-by-step solution

Given Information

Initial Velocity: \(v_0=200\mathrm{~ms}^{-1}\) Acceleration function: \(a(t)=3t^2-2t\) We use the following formula to find the velocity function, where \(C\) represents the integration constant: \[v(t)=\int a(t)\, dt + C\] #Step 2: Integrate the acceleration function#

Integration of Acceleration Function

Integrate the function \(a(t)\) with respect to \(t\): \[v(t)=\int(3t^2-2t) \, dt = t^3 - t^2+C\] #Step 3: Determine the integration constant#

Find Integration Constant

We know that the initial velocity is \(200 \mathrm{~ms}^{-1}\) at \(t=0\), we have: \[v(0)=0^3 - 0^2+C=200\] So, the integration constant \(C=200\). #Step 4: Write down the velocity function#

Velocity Function

Now that we have found the integration constant, we can write down the velocity function: \[v(t)=t^3 - t^2+200\] #Step 5: Find the velocity at 10 seconds#

Velocity at 10 Seconds

Plug in \(t=10\) into the velocity function to find the velocity at 10 seconds: \[v(10)=(10)^3 - (10)^2+200=1000-100+200=1100\mathrm{~ms}^{-1}\] The velocity of the particle at \(10\) seconds is \(1100\mathrm{~ms}^{-1}\). The correct answer is (A) \(1100\mathrm{~ms}^{-1}\).

Key concepts

Calculus in Physics

Calculus plays a fundamental role in understanding and solving problems in physics, particularly when dealing with motion and kinematics. It helps us to describe changes and solve problems involving rates of change. In physics, motion is often described by quantities like position, velocity, and acceleration. Each of these can vary with time.

For instance, acceleration can change as time progresses, and calculus provides the tools needed to determine how velocity changes in response to varying acceleration. Key aspects of calculus used in physics include:

• Derivatives: They help find rates, like how position changes with respect to time, which we know as velocity.
• Integrals: These are used to determine cumulative quantities, like total displacement over time from a velocity function.
• Differential Equations: They describe relationships involving rates of change and are essential in modeling real-world phenomena.

Understanding these concepts can allow a student to move from just knowing formulas to being able to derive and apply them to more complex, changing systems.

Velocity Calculation

Velocity represents how quickly an object's position changes with time. It's a vector quantity, meaning it has both magnitude and direction. Calculating velocity can be quite straightforward if the motion is simple, but what if the acceleration isn't constant? That's where calculus helps.

If we start with an initial velocity and have an acceleration function that varies with time, as shown in the original exercise, we need to find the velocity function first. This involves integration—essentially the opposite of finding a derivative.

In practical steps:

• Always begin with any given initial velocity. In this case, it was given as 200 m/s.
• Determine how velocity changes with time through the integration of the acceleration function.
• Substitute specific time values into the velocity function to find velocity at that moment.

Acquiring these skills is integral to tackling various kinematic problems in physics, particularly when dealing with non-constant accelerations.

Integration of Acceleration

Integration transforms an acceleration function, which describes the rate of change of velocity, into a velocity function. This fundamental calculus process involves computing the antiderivative. Here's how it works in simple terms:

Our acceleration function from the problem is given by:\[a(t) = 3t^2 - 2t\]
Integrating this function with respect to time gives us the velocity function:\[v(t) = \int (3t^2 - 2t) \, dt = t^3 - t^2 + C\]
Here, \(C\) is the constant of integration we find using initial conditions. At time \(t = 0\), velocity \(v(0)\) is 200 m/s, allowing us to solve for \(C\).

• The antiderivative of \(t^2\) is \(t^3/3\), and \(t\) turns into \(t^2/2\), illustrated simply here.
• Apply initial conditions to solve for constants, ensuring the resulting function accurately represents the system's initial state.

Thus, the velocity function translates a potentially complex physics problem into manageable math, providing insights into how a particle's motion evolves over time given its acceleration.