Problem 40 - Master Resource Book in JEE Main Physics

Question

A particle has an initial velocity of \(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\) and acceleration of \(0.4 \hat{\mathbf{i}}+0.3 \hat{\mathbf{j}}\). Its speed after \(10 \mathrm{~s}\) is (a) \(7 \sqrt{2}\) units (b) 7 units (c) \(8.5\) units (d) 10 units

Step-by-Step Solution

Answer

The speed after 10 seconds is (a) \(7 \sqrt{2}\) units.

Step 1: Identify Known Variables

We are given the initial velocity \( \mathbf{v}_0 = 3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} \) and the acceleration \( \mathbf{a} = 0.4 \hat{\mathbf{i}} + 0.3 \hat{\mathbf{j}} \). The time \( t \) is given as 10 seconds. We need to find the speed of the particle after 10 seconds.

Step 2: Calculate Final Velocity Components

The final velocity can be calculated using the formula \( \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t \). Compute the components of \( \mathbf{v} \):- For \( \hat{\mathbf{i}} \): \( v_{x} = 3 + 0.4 \times 10 = 3 + 4 = 7 \)- For \( \hat{\mathbf{j}} \): \( v_{y} = 4 + 0.3 \times 10 = 4 + 3 = 7 \)

Step 3: Determine Magnitude of Final Velocity

Find the magnitude of the final velocity \( \mathbf{v} = v_{x} \hat{\mathbf{i}} + v_{y} \hat{\mathbf{j}} \). The magnitude is given by the equation \( \| \mathbf{v} \| = \sqrt{v_{x}^2 + v_{y}^2} \):\[ \| \mathbf{v} \| = \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} = 7 \sqrt{2} \]

Step 4: Verify the Correct Answer Option

The calculated speed \( 7 \sqrt{2} \) matches option (a). Therefore, the correct answer is option (a).

Key Concepts

Understanding Initial Velocity

Initial velocity is the speed and direction a particle possesses when it begins its motion. In the exercise, the initial velocity of the particle is given as a vector: \( \mathbf{v}_0 = 3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} \). This tells us that the particle is moving in both the \( \hat{\mathbf{i}} \) (x-axis direction) and \( \hat{\mathbf{j}} \) (y-axis direction).

• The component \( 3 \hat{\mathbf{i}} \) represents the horizontal velocity.
• The component \( 4 \hat{\mathbf{j}} \) represents the vertical velocity.

Initial velocity is crucial in kinematics as it affects how the particle moves over time. Knowing this helps us predict how fast and in what direction the particle will continue its movement when considering other factors like acceleration.

The Role of Acceleration

Acceleration is the rate at which the velocity of an object changes with time. It adds a new dimension to the motion by affecting both speed and direction. In this problem, the particle's acceleration is given as \( \mathbf{a} = 0.4 \hat{\mathbf{i}} + 0.3 \hat{\mathbf{j}} \). This means the particle accelerates at different rates along the x-axis and y-axis.

• Acceleration of \( 0.4 \hat{\mathbf{i}} \) increases its horizontal speed over time.
• Acceleration of \( 0.3 \hat{\mathbf{j}} \) increases its vertical speed over time.

Acceleration allows the initial velocity to change, leading to a new final velocity after a certain period.

Calculating Final Velocity

The final velocity of an object is the velocity at a specific moment after it has been moving. In this scenario, it involves both the initial velocity and the effects of acceleration over time.
To compute it, we use the formula:
\( \mathbf{v} = \mathbf{v}_0 + \mathbf{a} t \).
After substituting the given values, we get:
- Horizontal component: \( v_{x} = 3 + 0.4 \times 10 = 7 \)
- Vertical component: \( v_{y} = 4 + 0.3 \times 10 = 7 \)
The resulting final velocity vector is \( \mathbf{v} = 7 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} \). This tells us that the particle now moves equally along both axes, with an enhanced velocity due to the given acceleration over the 10 seconds.

Understanding Velocity Magnitude

Velocity magnitude, also known as speed, is the length of the velocity vector. It reflects how fast an object is moving regardless of its direction. To find it, we calculate the magnitude of the final velocity vector using the formula: \( \| \mathbf{v} \| = \sqrt{v_{x}^2 + v_{y}^2} \). For our example:
\[ \| \mathbf{v} \| = \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} = 7 \sqrt{2} \]
This magnitude tells us the speed of the particle after the acceleration impacts are considered over the 10-second period. Here, it's crucial to note that while velocity is a vector and affected by both magnitude and direction, speed is merely the size of this vector, showing how rapidly the object is traveling.