/*! 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 37 A scooterist sees a bus \(1 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A scooterist sees a bus \(1 \mathrm{~km}\) ahead of him moving with a velocity of \(10 \mathrm{~ms}^{-1}\). With what speed the scooterist should move so as to overtake the bus in \(\begin{array}{ll}100 \mathrm{~s} . ? & \text { [Orissa JEE 2008] }\end{array}\) (a) \(10 \mathrm{~ms}^{-1}\) (b) \(20 \mathrm{~ms}^{-1}\) (c) \(50 \mathrm{~ms}^{-1}\) (d) \(30 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The scooterist should travel at \(20 \mathrm{~ms}^{-1}\). Option (b) is correct.

Step by step solution

01

Understand the Scenario

The problem involves a scooterist trying to overtake a bus that is 1 km ahead. The bus moves at a velocity of \(10 \mathrm{~ms}^{-1}\). The scooterist needs to calculate his speed to overtake the bus in 100 seconds.
02

Convert Distance to Meters

Since the distance is given in kilometers, we need to convert it to meters for consistency with the velocity units. Thus, \(1 \text{ km} = 1000 \text{ m}\).
03

Calculate the Distance Bus Will Travel

First, calculate the total distance the bus will cover in 100 seconds. The distance can be found using the formula: \( \text{Distance} = \text{Velocity} \times \text{Time}\). Thus, \(\text{Distance}_{\text{Bus}} = 10 \mathrm{~ms}^{-1} \times 100 \mathrm{~s} = 1000 \text{ m}\).
04

Determine Total Distance Scooterist Must Cover

To overtake the bus, the scooterist must cover the initial distance to the bus plus the distance the bus travels in 100 seconds: \(1000 \text{ m} + 1000 \text{ m} = 2000 \text{ m}\).
05

Find the Required Speed of Scooterist

Using the formula \( \text{Velocity} = \text{Distance}/\text{Time}\), calculate the velocity needed for the scooterist to cover 2000 m in 100 seconds. So, \( \text{Velocity}_{\text{Scooterist}} = 2000 \text{ m} / 100 \mathrm{~s} = 20 \mathrm{~ms}^{-1}\).
06

Select the Correct Answer from Options

After all calculations, the scooterist should travel at \(20 \mathrm{~ms}^{-1}\) to overtake the bus. Thus, the correct answer is (b) \(20 \mathrm{~ms}^{-1}\).

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Key Concepts

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

kinematics
Kinematics is the branch of physics that deals with motion without considering the forces that cause such motion. It is generally used to describe object movements using parameters like displacement, velocity, and acceleration. Understanding kinematics is crucial when solving relative velocity problems, like the one involving the scooterist and the bus. In this scenario, both the scooterist and the bus have velocities that determine how quickly they move toward or away from each other. This exercise requires calculating an unknown velocity using known distances and the time duration. By using kinematic equations, like \( \text{Velocity} = \frac{\text{Distance}}{\text{Time}} \), one can easily decipher the needed velocity for the scooterist to overtake the bus successfully. The exercise showcases the application of basic kinematics concepts to determine motion parameters required to meet a specific condition.
time and distance problems
Time and distance problems are classic exercises in physics and mathematics where you need to calculate one of the three key variables: time, distance, and speed (or velocity). Such problems often use direct relationships like \( \text{Distance} = \text{Velocity} \times \text{Time} \), and they require conversions for consistency in units. In the original problem, the scooterist must overcome a relative distance of 2000 meters within 100 seconds to overtake the bus. These types of problems test your ability to apply logical reasoning and manage units correctly – like converting kilometers to meters for uniform calculations. Time and distance questions sharpen problem-solving skills by requiring an understanding of relative motion through time, which is especially useful in competitive exams.
Orissa JEE Physics 2008
Orissa JEE Physics 2008 is a state-level examination that assesses understanding in various physics concepts, often through practical problem-solving exercises. This particular question from the 2008 exam challenges students to apply their understanding of kinematics and relative velocity. By presenting a real-world scenario where a scooterist attempts to overtake a bus, this problem helps to test theoretical knowledge and its practical application. Importantly, it encourages efficiency in calculating time and distance, which are integral in situations involving moving objects. The exam tests not only students' grasp of physics concepts but also their ability to navigate and solve real-life physics-based problems.

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Most popular questions from this chapter

The displacement of a body along \(x\)-axis depends on time as \(\sqrt{x}=t+1\). Then, the velocity of body (a) increase with time (b) decrease with time (c) independent of time (d) None of these

On a two lane road, car \(A\) is travelling with a speed of \(36 \mathrm{~km} / \mathrm{h}\). Two cars \(B\) and \(C\) approach car \(A\) in opposite directions with a speed of \(54 \mathrm{~km} / \mathrm{h}\) each. At a certain instant, when the distance \(A B\) is equal to \(A C\), both being \(1 \mathrm{~km}, B\) decides to overtake \(A\) before \(C\) does. In this case, the acceleration of car \(B\) is required to avoid an accident (a) \(1 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(0.1 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(1.9 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(0.2 \mathrm{~m} / \mathrm{s}^{2}\)

A particle is moving with a uniform acceleration along a straight line \(A B\). Its speed at \(A\) and \(B\) are \(2 \mathrm{~ms}^{-1}\) and \(14 \mathrm{~ms}^{-1}\) respectively. Then (a) its speed at mid-point of \(A B\) is \(10 \mathrm{~ms}^{-1}\) (b) its speed at a point \(P\) such that \(A P: P B=1: 5\) is \(4 \mathrm{~ms}^{-1}\) (c) the time to go from \(A\) to mid-point of \(A B\) is double of that to go from mid-point to \(B\) (d) None of the above

The motion of a body is given by the equation \(\frac{d v(t)}{d t}=6.0-3 v(t)\), where \(v(t)\) is speed in \(\mathrm{ms}^{-1}\) and \(t\) in second. If body was at rest at \(t=0\) (a) the terminal speed is \(2.0 \mathrm{~ms}^{-1}\) (b) the speed varies with the times as \(v(t)=2\left(1-e^{-3 t}\right) \mathrm{ms}^{-1}\) (c) the speed is \(1.0 \mathrm{~ms}^{-1}\) when the acceleration is half the initial value (d) the magnitude of the initial acceleration is \(6.0 \mathrm{~ms}^{-2}\)

The driver of a car moving with a speed of \(10 \mathrm{~ms}^{-1}\) sees a red light ahead, applies brakes and stops after covering \(10 \mathrm{~m}\) distance. If the same car were moving with a speed of \(20 \mathrm{~ms}^{-1}\), the same driver would have stopped the car after covering \(30 \mathrm{~m}\) distance. Within what distance the car can be stopped if travelling with a velocity of \(15 \mathrm{~ms}^{-1}\) ? Assume the same reaction time and the same deceleration in each case. (a) \(18.75 \mathrm{~m}\) (b) \(20.75 \mathrm{~m}\) (c) \(22.75 \mathrm{~m}\) (d) \(25 \mathrm{~m}\)

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