/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 A bus moves over a straight leve... [FREE SOLUTION] | 91Ó°ÊÓ

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A bus moves over a straight level road with a constant acceleration \(a .\) A body in the bus drops a ball outside. The acceleration of the ball with respect to the bus and the earth are respectively (a) \(a\) and \(g\) (b) \(a+g\) and \(g-a\) (c) \(\sqrt{a^{2}+g^{2}}\) and \(g\) (d) \(\sqrt{a^{2}+g^{2}}\) and \(a\)

Short Answer

Expert verified
(c) \(\sqrt{a^2 + g^2}\) and \(g\).

Step by step solution

01

Understand the Scenario

In this problem, we have a bus moving with a constant acceleration \(a\) along a straight road. A ball is dropped from the bus. We need to determine the acceleration of the ball both with respect to the bus and the earth.
02

Determine Acceleration with Respect to the Earth

When the ball is dropped from the bus, its horizontal motion continues with the bus's velocity at the instant of release. However, vertically, it is only influenced by gravity \(g\). Therefore, the acceleration of the ball with respect to the earth is \(g\) downwards, as gravity is the only force acting on it in the vertical direction.
03

Determine Acceleration with Respect to the Bus

From the viewpoint of the bus, which is accelerating forward with acceleration \(a\), the ball does not experience any horizontal acceleration because it does not continue accelerating after being dropped. Thus, it appears to have an acceleration of \(-a\) relative to the bus in the horizontal direction. Vertically, it accelerates downwards at \(g\) due to gravity. Using the Pythagorean theorem to combine these two perpendicular accelerations, the magnitude of acceleration with respect to the bus is \(\sqrt{a^2 + g^2}\).
04

Compare with Options

With the analysis above, we identify that the acceleration of the ball with respect to the bus is \(\sqrt{a^2 + g^2}\), and with respect to the earth, it is \(g\). Comparing this with the given options, the correct answer is option (c) \(\sqrt{a^2 + g^2}\) and \(g\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Projectile Motion
Projectile motion is a type of motion experienced by an object that is projected into the air and moves under the influence of gravity alone. It's crucial to note that in projectile motion, there is no acceleration in the horizontal direction once the object is in motion, assuming air resistance is negligible.

The only force acting on the projectile is gravity, which results in a constant acceleration downwards. This vertical acceleration due to gravity is represented as \(g\). A key aspect of projectile motion is that horizontal and vertical motions are independent of each other. This means you can analyze them separately before combining them for a complete picture of the trajectory.

Some fundamental characteristics of projectile motion include:
  • The initial horizontal velocity of the object will remain constant throughout the motion because no external forces (other than air resistance, which is often neglected) are acting in the horizontal direction.
  • The motion vertically is uniformly accelerated due to gravity, which pulls the object downward at \(9.8 \, m/s^2\) on Earth.
In the context of the exercise, when the ball is dropped from the moving bus, it continues to move horizontally with the bus's velocity while accelerating downward due to gravity.
Constant Acceleration
Constant acceleration refers to the condition wherein an object's rate of velocity change does not vary over time. This is a fundamental concept in mechanics, simplifying the analysis of motion because equations become linear.

When an object moves with constant acceleration, you can use the following kinematic equations to describe its motion:
  • Final velocity, \(v = u + a t\), where \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.
  • Displacement, \(s = ut + \frac{1}{2}at^2\).
  • Final velocity squared, \(v^2 = u^2 + 2as\).
For the ball in our problem, while it moves with the bus, it shares the same constant acceleration \(a\) horizontally but after being dropped; the acceleration that needs consideration is the gravitational force acting vertically. The bus continues to accelerate at \(a\) while the ball only accelerates downwards due to gravity (\(g\)). By understanding this distinction, you can explain the different accelerations relative to the bus and the Earth.
Newton's Laws of Motion
Newton's laws of motion provide the foundation for understanding the relationship between motion and forces acting on an object. Each law explains a fundamental aspect of motion under the influence of forces.

Newton’s First Law states that an object at rest stays at rest and an object in motion stays in motion at a constant velocity unless acted upon by an external force. In the exercise, the ball continues moving horizontally with the bus's velocity because no horizontal force acts on it after being dropped.

Newton's Second Law introduces the fundamental concept of force, expressed as \(F = ma\). This law underscores how the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. When applied to our scenario, it explains why the ball continues its horizontal velocity (because no net force acts horizontally) and accelerates vertically under gravity alone.

Newton’s Third Law posits that for every action, there is an equal and opposite reaction. While it's less prominently featured in the ball’s motion, understanding this law can help comprehend interactions within the system, such as the reaction forces between the ball and the air it moves through.

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Most popular questions from this chapter

A particle covers \(4 \mathrm{~m}, 5 \mathrm{~m}, 6 \mathrm{~m}\) and \(7 \mathrm{~m}\) in \(3 \mathrm{rd}, 4\) th, 5th and 6th second respectively. The particle starts (a) with an initial non-zero velcoity and moves with uniform acceleration (b) from rest and moves with uniform velocity (c) with an initial velocity and moves with uniform velcoity (d) from rest and moves with uniform acceleration

Two trains travelling on the same track are approaching each other with equal speeds of \(40 \mathrm{~ms}^{-1}\). The drivers of the trains begin to decelerate simultaneously when they are just \(2 \mathrm{~km}\) apart. If the decelerations are both uniform and equal, then the value of deceleration to barely avoid collision should be (a) \(0.8 \mathrm{~ms}^{-2}\) (b) \(2.1 \mathrm{~ms}^{-2}\) (c) \(11.0 \mathrm{~ms}^{-2}\) (d) \(13.2 \mathrm{~ms}^{-2}\)

Two trains \(A\) and \(B\) of length \(400 \mathrm{~m}\) each are moving on two parallel tracks with a uniform speed of \(72 \mathrm{~km} / \mathrm{h}\) in the same direction, with \(A\) ahead of \(B\). The driver of \(B\) decides to overtake \(A\) and accelerates by \(1 \mathrm{~m} / \mathrm{s}^{2}\). If after \(50 \mathrm{~s}\), the guard of \(B\) just brushes past the driver of \(A\), what was the original distance between them? (a) \(1250 \mathrm{~m}\) (b) \(1350 \mathrm{~m}\) (c) \(1450 \mathrm{~m}\) (d) None of these

The displacement-time graphs of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(x\)-axis. The ratio of the two velocities is (a) \(\sqrt{3}: 1\) (b) \(1: 1\) (c) \(1: 2\) (d) \(1: \sqrt{3}\)

From the top of a tower of height \(50 \mathrm{~m}\), a ball is thrown vertically upwards with a certain velocity. It hits the ground \(10 \mathrm{~s}\) after it is thrown up. How much time does it take to cover a distance \(A B\) where \(A\) and \(B\) are two points \(20 \mathrm{~m}\) and \(40 \mathrm{~m}\) below the edge of the tower? \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(2.0 \mathrm{~s}\) (b) \(1.0 \mathrm{~s}\) (c) \(0.5 \mathrm{~s}\) (d) \(0.4 \mathrm{~s}\)

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