Question

Match column type question At any instant, the horizontal position to balloon is defined by \(\mathrm{x}=\mathrm{gt} \mathrm{m} .\) Equation of path is \(\left(\mathrm{x}^{2} / 30\right)\). After \(\mathrm{t}=2 \mathrm{sec}\), match the following. Table 1 Table 2 (A) distance of balloon from station \(\mathrm{A}\). (P) \(14.1,50^{\circ}\) (B) magnitude \& direction of velocity (Q) \(24 \mathrm{~m}\) C) magnitude \& direction of acceleration (R) \(21 \mathrm{~m}\) (S) \(5.4,90^{\circ}\) (A) \(\mathrm{A}-\mathrm{R}, \mathrm{B}-\mathrm{S}, \mathrm{C}-\mathrm{P}\) (B) \(\mathrm{A}-\mathrm{Q}, \mathrm{B}-\mathrm{S}, \mathrm{C}-\mathrm{P}\) (C) \(\mathrm{A}-\mathrm{R}, \mathrm{B}-\mathrm{P}, \mathrm{C}-\mathrm{S}\) (D) \(\mathrm{A}-\mathrm{Q}, \mathrm{B}-\mathrm{P}, \mathrm{C}-\mathrm{S}\)

Short answer

The correct answer is (D): A-Q, B-P, C-S.

Step-by-step solution

Find the horizontal position x

At any instant, the horizontal position x of the balloon is described by the equation \(x = gtm\). Given that \(t = 2s\), with \(g = 10 m/s^2\), we have: \[x = 10 \times 2m = 20m\]

Find the vertical position y

To obtain the vertical position of the balloon at \(t = 2s\), we use the given equation for the path of the balloon: \[y = \frac{x^2}{30}\] Substitute the value of x found in Step 1: \[y = \frac{(20)^2}{30} = \frac{400}{30} = 13.33m\]

Determine the distance of the balloon from the station

Now, we compute the distance (A) of the balloon from the station A using the Pythagorean theorem: \(A = \sqrt{x^2 + y^2}\) Substitute the values of x and y found in Steps 1 and 2: \[A = \sqrt{(20)^2 + (13.33)^2} = \sqrt{400 + 177.77} ≈ 24m\] Verify the correct match from Table 1 and 2: A matches with Q.

Compute the magnitude and direction of the velocity (v_x and v_y)

Given \(x = gtm\), we have: \[\frac{dx}{dt} = v_x = gt\] The horizontal velocity \(v_x = 10 \times 2 = 20m/s\) Since the balloon moves vertically following a parabolic path, we can differentiate the equation \(y = \frac{x^2}{30}\) with respect to time t: \[\frac{dy}{dt} = \frac{d}{dt} [\frac{x^2}{30}]\] Since \(\frac{dx}{dt} = v_x\), we get: \[\frac{dy}{dt} = \frac{2x \times v_x}{30}\] Substituting the values of x and \(v_x\), we get: \[\frac{dy}{dt} = v_y = \frac{2 \times 20 \times 20}{30} = 26.67 m/s\] Now, compute the magnitude of the velocity (B): \[v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20)^2 + (26.67)^2} ≈ 33.33 m/s\] Next, find the direction of the velocity in degrees: \[\theta = \tan^{-1}(\frac{v_y}{v_x}) = \tan^{-1}(\frac{26.67}{20}) ≈ 53.33^{\circ}\] Verify the correct match from Table 1 and 2: B matches with P.

Calculate the magnitude and direction of the acceleration (a_x and a_y)

In the horizontal direction, \(a_x = 0\) (constant horizontal velocity) In the vertical direction, we second differentiate the equation \(y = \frac{x^2}{30}\): \[\frac{d^2y}{dt^2} = \frac{d}{dt} [\frac{2x \times v_x}{30}]\] Since \(a_x = 0\), we have: \[\frac{d^2y}{dt^2} = a_y = \frac{2 \times v_x^2}{30}\] Substituting the value of \(v_x\): \[a_y = \frac{2 \times 20^2}{30} = 26.67 m/s^2\] Compute the magnitude of the acceleration (C): \[a = \sqrt{a_x^2 + a_y^2} = \sqrt{(0)^2 + (26.67)^2} = 26.67m/s^2\] Then, find the direction of the acceleration in degrees: \[\alpha = \tan^{-1}(\frac{a_y}{a_x}) \] Since \(a_x = 0\), we have: \[\alpha = 90^{\circ}\] Verify the correct match from Table 1 and 2: C matches with S. Based on our calculations and matches: The correct answer is (D): A-Q, B-P, C-S.

Key concepts

Kinematics

Kinematics is the study of motion without considering the forces that cause it. It focuses on describing how objects move, their speed, direction, and the path of their movement. In simple terms, kinematics is all about understanding the "how" of motion, not the "why."

For our balloon example, kinematics helps us determine its position and velocity at a given time. We describe its horizontal position as a function of time using the equation \( x = gt \), where \( g \) is the acceleration due to gravity and \( t \) is the time. This means the balloon's horizontal position increases at a constant rate over time.

To find its vertical position, we used the equation \( y = \frac{x^2}{30} \), which depicts a parabolic path. This indicates how vertically the balloon rises with respect to its horizontal movement. Knowing these kinematic descriptions allows us to further compute the object's velocity and acceleration in the context of motion.

Equations of Motion

Equations of motion form the mathematical backbone of kinematics, offering precise calculations of an object's location, speed, and trajectory. These equations connect initial velocity, acceleration, time, and displacement, making it possible to predict future movement.

For the problem at hand, the horizontal motion is simplified into the equation \( x = gt \), which straightforwardly relates time to position with a known acceleration. This equation is crucial as it allows us to identify the horizontal component of motion independent of vertical movement.

The vertical motion, however, requires understanding the parabolic path described by \( y = \frac{x^2}{30} \). This equation involves squaring the horizontal position to determine how the height changes, reflecting the nature of projectile motion where gravity influences the vertical motion, creating a curved path. By using these equations, one can precisely calculate other essential motion characteristics like velocity and acceleration.

Velocity and Acceleration

Velocity and acceleration are fundamental aspects of motion, describing how fast something moves and how its speed changes over time.

In our balloon problem:

• **Velocity**: Velocity has both magnitude and direction. We derived its horizontal component \( v_x = gt \), thus \( v_x = 20 \) m/s. For the vertical component, \( v_y \) was found to be 26.67 m/s. These together give a full velocity magnitude \( v = \sqrt{(20)^2 + (26.67)^2} \) and a direction \( \theta = \tan^{-1}(\frac{v_y}{v_x}) \approx 53.33^{\circ} \).


• **Acceleration**: In this example, horizontal acceleration \( a_x \) is zero, indicating constant horizontal velocity. Vertically, assuming a parabolic path affected by gravity, the acceleration \( a_y \) leads to \( a = \sqrt{0^2 + (26.67)^2} \), which maintains a vertical direction of \( 90^{\circ} \), due to gravity.

This breakdown of velocity and acceleration not only predicts where and how fast the balloon moves but also how its speed changes, providing a complete picture of its motion.