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During the shooting of a super hit film 'MARD' Amitabh Bachchan was waiting for his beloved 'Amrita Singh' with his dog. When he saw her approaching, the dog was excited and dashed to her then back to master and so on, never stopping. How far would you estimate the dog ran if its speed is \(30 \mathrm{~km} / \mathrm{hr}\) and each of them walked at \(4 \mathrm{~km} / \mathrm{hr}\), starting from a distance \(400 \mathrm{~m}\) apart? (a) \(400 \mathrm{~m}\) (b) \(880 \mathrm{~m}\) (c) \(1500 \mathrm{~m}\) (d) \(30 \mathrm{~km}\)

Short Answer

Expert verified
The dog ran 1500 meters (c).

Step by step solution

01

Understand the Movement

The dog runs back and forth between Amitabh Bachchan and Amrita Singh without stopping, maintaining a constant speed, while the two walk towards each other at a constant speed.
02

Calculate Time Taken to Meet

The total distance between Amitabh and Amrita is given as 400 meters or 0.4 kilometers. They move towards each other with a combined speed of \(4 + 4 = 8 \, \mathrm{km/hr}\). Thus, the time taken to meet is given by: \[ t = \frac{\text{Distance}}{\text{Speed}} = \frac{0.4 \, \mathrm{km}}{8 \, \mathrm{km/hr}} = 0.05 \, \mathrm{hr} \]
03

Calculate the Distance Ran by the Dog

Given the dog's speed is \(30 \, \mathrm{km/hr}\), the total distance the dog runs while they meet is determined by multiplying the dog's speed by the time taken for them to meet:\[ \text{Distance} = 30 \, \mathrm{km/hr} \times 0.05 \, \mathrm{hr} = 1.5 \, \mathrm{km} \]
04

Convert Distance to Meters

Since the answer choices are given in meters, convert the distance the dog ran from kilometers to meters:\[ 1.5 \, \mathrm{km} = 1500 \, \mathrm{m} \]

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Key Concepts

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

Kinematics Unveiled
In physics, particularly in kinematics, we study motion without considering the forces that cause it. It is all about describing the movement of objects through their velocity and acceleration. Kinematics often involves breaking down complex motion into simpler elements like speed, direction, and time. For instance, in our exercise, the dog exhibits simple back-and-forth motion between two people. This scenario perfectly illustrates how kinematics can predict the dog's behavior through understanding its constant speed and the time involved.
Because the dog's speed is constant at 30 km/hr, our kinematic calculations focus on how far it can travel within the timeframe that the two people meet. The dog's repeated motion doesn't complicate the calculation since it maintains the same speed, helping us calculate accurately using Newton's equations.
Distance-Time Relationship Explained
The distance-time relationship is a crucial concept in understanding motion. This relationship states that distance covered is equal to speed multiplied by time. This is a straightforward yet powerful equation that allows us to analyze and predict the outcome of moving objects.
In the exercise, the total distance between Amitabh and Amrita is 400 meters. They walk towards each other, closing in on this distance. By knowing both their total speed of approaching each other, which is 8 km/hr, and the initial distance, we can find the time it takes for them to meet. Once we have the time, we multiply it by the dog's speed to determine the distance it runs while maintaining the same pace during the entire period.
  • The formula is: \[ \text{Distance} = \text{Speed} \times \text{Time} \]
  • This allows us to use a single formula to determine different aspects of motion.
  • Understanding this relationship simplifies complex motion studies into manageable calculations.
Understanding Unit Conversion
Unit conversion is an essential skill for solving real-world problems in physics. Often, the measurements we obtain or need to use are in different units and require conversion to make sense in calculations.
In our exercise, the distance unit differs in the problem setup (400 meters) and the calculations (in kilometers). Calculations in physics typically obey the metric system, but results often need conversions to match answer choices, as seen here. Converting kilometers to meters involves multiplying by 1000:
  • For example, \(1.5\, \text{km} = 1500\, \text{m}\).
  • Thus, understanding the basic conversion between meters and kilometers helps ensure your final answers are in the correct unit, preventing unnecessary mistakes.
Mastering unit conversion not only aids in solving physics problems but is a skill that enhances logical reasoning and quantitative evaluation in various scientific fields.

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Most popular questions from this chapter

A body starts from rest and moves with a constant acceleration. The ratio of distance covered in the \(n\) th second to the distance covered in \(n\) second is : (a) \(\frac{2}{n}-\frac{1}{n^{2}}\) (b) \(\frac{1}{n^{2}}-\frac{1}{n}\) (c) \(\frac{2}{n^{2}}-\frac{1}{n}\) (d) \(\frac{2}{n}+\frac{1}{n^{2}}\)

One rickshaw leaves Patna Junction for Gandhi Maidan at every 10 minute. The distance between Gandhi Maidan and Patna Junction is \(6 \mathrm{~km}\). The rickshaw travels at the speed of \(6 \mathrm{~km} / \mathrm{hr}\). What is the number of rickshaw that a rickshaw puller driving from Gandhi Maidan to Patna Junction must be in the route if he starts from Gandhi Maidan simultaneously with one of the rickshaw leaving Patna Junction: (a) 11 (b) 12 (c) 5 (d) 1

To a person going toward east in a car with a velocity of \(25 \mathrm{~km} / \mathrm{hr}\), a train appears to move towards north with a velocity of \(25 \sqrt{3} \mathrm{~km} / \mathrm{hr}\). The actual velocity of the train will be : (a) \(25 \mathrm{~km} / \mathrm{hr}\) (b) \(50 \mathrm{~km} / \mathrm{hr}\) (c) \(5 \mathrm{~km} / \mathrm{hr}\) (d) \(53 \mathrm{~km} / \mathrm{hr}\)

Balls are thrown vertically upward in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is \(5 \mathrm{~m}\), the number of balls thrown per minute will be: (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) 60 (b) 40 (c) 50 (d) 120

A ball is dropped vertically from a height \(d\) above the ground. It hits the ground and bounces up vertically to a height \(d / 2 .\) Neglecting subsequent motion and air resistance, its speed \(v\) varies with the height \(h\) above the ground as :

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