Problem 43 - Master Resource Book in JEE Main Physics

Question

A body moves with initial velocity \(10 \mathrm{~ms}^{-1}\). If it covers a distance of \(20 \mathrm{~m}\) in \(2 \mathrm{~s}\) then acceleration of the body is [Orissa JEE 2011] (a) zero (b) \(10 \mathrm{~ms}^{-2}\) (c) \(5 \mathrm{~ms}^{-2}\) (d) \(2 \mathrm{~ms}^{-2}\)

Step-by-Step Solution

Answer

The acceleration of the body is (a) zero.

Step 1: Identify The Known Values

We know the initial velocity \( u \) is \( 10 \ \mathrm{ms}^{-1} \), the distance \( s \) covered is \( 20 \ \mathrm{m} \), and the time \( t \) taken is \( 2 \ \mathrm{s} \).

Step 2: Select the Appropriate Equation

Use the equation of motion \( s = ut + \frac{1}{2} a t^2 \) to solve for acceleration \( a \).

Step 3: Substitute Known Values into the Equation

Substitute \( s = 20 \ \mathrm{m} \), \( u = 10 \ \mathrm{ms}^{-1} \), and \( t = 2 \ \mathrm{s} \) into the equation: \[ 20 = 10 \cdot 2 + \frac{1}{2} a \cdot (2)^2 \]

Step 4: Simplify the Equation

Simplify the right side of the equation: \[ 20 = 20 + 2a \]

Step 5: Solve for Acceleration

Rearrange the equation to solve for \( a \): \[ 0 = 2a \] \[ a = 0 \]

Step 6: Conclusion

The acceleration of the body is \( 0 \ \mathrm{ms}^{-2} \).

Key Concepts

Equations of Motion

Equations of motion are critical in understanding how objects move. They provide a mathematical framework to describe the motion of a body under uniform acceleration. These equations relate attributes like initial velocity, final velocity, displacement, acceleration, and time. It's important for solving problems involving motion in a straight line.

There are three primary equations of motion for constant acceleration:

• First: \( v = u + at \), where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time.
• Second: \( s = ut + \frac{1}{2} at^2 \), where \( s \) is displacement.
• Third: \( v^2 = u^2 + 2as \).

In our exercise, the second equation was used to find the acceleration of the body. This equation directly links displacement with initial velocity, time, and acceleration. By substituting the known values, we can solve for the unknown quantity. This approach is particularly useful when acceleration is constant

These equations also assume that the direction remains constant, making them ideal for vertical and horizontal motion under gravity.

Initial Velocity

Initial velocity is the speed at which an object begins its motion. It is a critical parameter in the equations of motion, particularly when calculating how far and how fast an object will travel over time. In simple terms, it's the starting speed of a moving object.

In the example given, the initial velocity \( u \) was provided as \( 10 \mathrm{~ms}^{-1} \). This is used in the equation \( s = ut + \frac{1}{2} at^2 \) to determine how the object's speed and distance traversed interact with its acceleration over time.

Initial velocity can be positive or negative depending on the direction of motion. This value is crucial for setting the baseline from which the object's future motion is calculated. It allows one to determine subsequent changes in speed and position when an object is subject to acceleration.

Acceleration

Acceleration is the rate at which an object's velocity changes with time. It can be thought of as how quickly an object speeds up or slows down. Positive acceleration indicates an increase in speed, while negative acceleration (often called deceleration) suggests a decrease.

In our exercise, the task was to find the acceleration given certain known values, including initial velocity and displacement. By applying the equation \( s = ut + \frac{1}{2} at^2 \), we found that the acceleration was \( 0 \mathrm{~ms}^{-2} \). This indicates uniform motion, meaning the object moved at a constant velocity without speeding up or slowing down.

Understanding acceleration is fundamental in kinematics since it impacts how objects travel over time. It explains changes in motion behavior, linking directly with forces applied to bodies, as per Newton's laws.

Displacement

Displacement refers to the change in position of an object. It's a vector quantity, meaning it has both magnitude and direction. Unlike distance, which is a scalar and only measures how far an object has traveled, displacement considers the straight-line path between the starting and ending points.

In the given problem, the displacement \( s \) was \( 20 \mathrm{~m} \), which represents the net change in position of the body. The equation of motion \( s = ut + \frac{1}{2} at^2 \) was used to incorporate displacement in solving for acceleration, showing how displacement affects the dynamic behavior of an object over time.

Displacement is crucial in physics because it helps define the specifics of an object's trajectory. By understanding displacement, one can better predict and calculate other motion parameters.