Problem 46 - 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) 10 units (b) \(7 \sqrt{2}\) units (c) 7 units (d) \(8.5\) units

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We need to find the speed of the particle after 10 seconds, given its initial velocity and constant acceleration. The speed is the magnitude of the velocity vector at that time.

Step 2: Identify Initial and Final Components

The initial velocity of the particle is given by the vector \( \mathbf{u} = 3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}} \). The acceleration is \( \mathbf{a} = 0.4 \hat{\mathbf{i}} + 0.3 \hat{\mathbf{j}} \). The time considered is 10 seconds. We need to calculate the final velocity using these values.

Step 3: Calculate Final Velocity

Use the kinematic equation for velocity: \( \mathbf{v} = \mathbf{u} + \mathbf{a}t \). Here, \( t = 10 \). Thus, \( \mathbf{v} = (3 + 0.4 \times 10) \hat{\mathbf{i}} + (4 + 0.3 \times 10) \hat{\mathbf{j}} \). Simplifying, we get \( \mathbf{v} = 7 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} \).

Step 4: Calculate Speed from Velocity

The speed of the particle is the magnitude of the final velocity vector \( \mathbf{v} = 7 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} \). The magnitude is \( |\mathbf{v}| = \sqrt{(7)^2 + (7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \).

Key Concepts

Velocity

Velocity represents both the speed and direction of a moving object. It is a vector quantity, which means it possesses both magnitude and direction.

Consider the initial velocity of our particle, given as \(3 \hat{\mathbf{i}} + 4 \hat{\mathbf{j}}\). In this expression:

• The term \(3 \hat{\mathbf{i}}\) represents the component of velocity in the x-direction.
• The term \(4 \hat{\mathbf{j}}\) represents the component of velocity in the y-direction.

When we combine these components, we get the overall velocity vector.

This vector nature allows us to easily calculate the final velocity when acceleration is involved. Using the kinematic equation \( \mathbf{v} = \mathbf{u} + \mathbf{a}t \), we add the effect of acceleration over time to the initial velocity to derive the new velocity.

For example, at \( t = 10 \) seconds, the final velocity becomes \( 7 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} \). This means, after 10 seconds, our particle is moving with equal speed in both x and y directions.

Acceleration

Acceleration describes the rate at which an object's velocity changes over time. It's also a vector quantity, meaning it has both magnitude and direction.

In the exercise, the particle's acceleration is given as \(0.4 \hat{\mathbf{i}} + 0.3 \hat{\mathbf{j}}\). This indicates:

• The object experiences a change in velocity of \(0.4\) units in the x-direction every second.
• The object experiences a change in velocity of \(0.3\) units in the y-direction every second.

Acceleration can influence both the speed and direction of motion. When we multiply acceleration by time (10 seconds in this case), we observe its cumulative effect on the initial velocity.

This is outlined in the equation \( \mathbf{v} = \mathbf{u} + \mathbf{a}t \), leading to a final velocity of \(7 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}}\). By understanding acceleration's impact, we can predict how the particle's velocity will evolve over a period of time.

Speed calculation

Calculating speed is about finding the magnitude of the velocity vector. Speed is a scalar quantity, meaning it only has magnitude - no directional component.

To find the speed after 10 seconds, we determine the magnitude of the final velocity vector \( \mathbf{v} = 7 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} \). The formula for the magnitude (or speed) is:

• \( |\mathbf{v}| = \sqrt{(v_x)^2 + (v_y)^2} \)

Here, \(v_x\) and \(v_y\) are the components of the velocity vector in the x and y directions respectively.
Substituting the values, we get:
\( |\mathbf{v}| = \sqrt{(7)^2 + (7)^2} = \sqrt{49 + 49} = \sqrt{98} \)
Thus, the speed is \( 7\sqrt{2} \).

It is important to note that while velocity tells us how fast and in what direction, speed only provides how fast the particle is moving. This understanding helps in distinguishing between the directional path and the rate of movement, which are both critical in kinematics problems.