• NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • Class 6-10
      • Class 6th
      • Class 7th
      • Class 8th
      • Class 9th
      • Class 10th
    • View All Options
      • Online Courses
      • Offline Courses
      • Distance Learning
      • Hindi Medium Courses
      • International Olympiad
    • NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE (Main+Advanced)
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE Main
      • Class 11th
      • Class 12th
      • Class 12th Plus
  • NEW
    • JEE 2025
    • NEET
      • 2024
      • 2023
      • 2022
    • Class 6-10
    • JEE Main
      • Previous Year Papers
      • Sample Papers
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • JEE Advanced
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • NEET
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • NCERT Solutions
      • Class 6
      • Class 7
      • Class 8
      • Class 9
      • Class 10
      • Class 11
      • Class 12
    • CBSE
      • Notes
      • Sample Papers
      • Question Papers
    • Olympiad
      • NSO
      • IMO
      • NMTC
    • TALLENTEX
    • AOSAT
    • ALLEN e-Store
    • ALLEN for Schools
    • About ALLEN
    • Blogs
    • News
    • Careers
    • Request a call back
    • Book home demo
Home
Class 12
PHYSICS
A man is at a distance of 6 m from a bus...

A man is at a distance of 6 m from a bus. The bus begins to move with a constant acceleration of `3 ms^(−2)`. In order to catch the bus, the minimum speed with which the man should run towards the bus is

To view this video, please enable JavaScript and consider upgrading to a web browser thatsupports HTML5 video

Doubtnut Promotions Banner Desktop LightDoubtnut Promotions Banner Desktop DarkDoubtnut Promotions Banner Mobile LightDoubtnut Promotions Banner Mobile Dark

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the minimum speed at which a man must run towards a bus that is accelerating away from him. Here’s a step-by-step breakdown of the solution: ### Step 1: Understand the Problem The man is initially 6 meters away from the bus, which starts moving with a constant acceleration of \( a = 3 \, \text{m/s}^2 \). We need to find the minimum speed \( v \) at which the man must run to catch the bus. ### Step 2: Set Up the Equations 1. **Distance covered by the bus**: The bus starts from rest and accelerates. The distance \( s_2 \) covered by the bus in time \( t \) is given by the equation: \[ s_2 = \frac{1}{2} a t^2 \] Substituting \( a = 3 \, \text{m/s}^2 \): \[ s_2 = \frac{1}{2} \cdot 3 \cdot t^2 = \frac{3}{2} t^2 \] 2. **Distance covered by the man**: The man runs towards the bus with speed \( v \). The distance \( s_1 \) he covers in time \( t \) is: \[ s_1 = v t \] ### Step 3: Relate the Distances To catch the bus, the total distance covered by the man must equal the initial distance plus the distance the bus has moved: \[ s_1 = 6 + s_2 \] Substituting the expressions for \( s_1 \) and \( s_2 \): \[ v t = 6 + \frac{3}{2} t^2 \] ### Step 4: Rearrange the Equation Rearranging the equation gives us: \[ \frac{3}{2} t^2 - v t + 6 = 0 \] This is a quadratic equation in \( t \). ### Step 5: Apply the Quadratic Formula For the quadratic equation \( at^2 + bt + c = 0 \), the solutions for \( t \) are given by: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = \frac{3}{2} \), \( b = -v \), and \( c = 6 \): \[ t = \frac{v \pm \sqrt{v^2 - 4 \cdot \frac{3}{2} \cdot 6}}{2 \cdot \frac{3}{2}} \] This simplifies to: \[ t = \frac{v \pm \sqrt{v^2 - 36}}{3} \] ### Step 6: Ensure Real Solutions For \( t \) to be real, the discriminant must be non-negative: \[ v^2 - 36 \geq 0 \] This implies: \[ v^2 \geq 36 \] Thus: \[ v \geq 6 \] ### Step 7: Conclusion The minimum speed \( v \) that the man must run to catch the bus is: \[ v = 6 \, \text{m/s} \] Thus, the answer is **6 meters per second**.

Topper's Solved these Questions

  • AIIMS 2015

    AIIMS PREVIOUS YEAR PAPERS|Exercise PHYSICS|60 Videos
  • AIIMS 2017

    AIIMS PREVIOUS YEAR PAPERS|Exercise PHYSICS|59 Videos

Similar Questions

Explore conceptually related problems

A man is 45 m behind the bus when the bus starts acceleration from rest with acceleration 2.5(m)/(s^2) . With what minimum velocity should man start running to catch the bus?

A passenger is standing `d` metres away from a bus. The bus begins to move eith constat acceleration `a. To catch the bus the passenger runs at a constant speed (v) towards the bus, What must so that he may catch the bus.

A bus begins to move with an acceleration of 1 ms^(-1) . A man who is 1 48m behind the bus starts running at 10 ms^(-1) to catch the bus, the man will be able to catch the bus after .

A man is d distance behind the bus when the bus starts accelerating from rest with an acceleration a_(0) . With what minimum constant velocity should the man start running to catch the bus

A person is standing at a distance 's' m from a bus. The bus begins to move with cosntant acceleration 'a' m//s^(2) away from the person. To catch the bus, the person runs at a constant speed 'v' m/s towards the bus. Minimum speed of the person so that he can catch the bus is:

A passenger is standing d meters away from a bus. The bus beings to move with constant acceleration a. To catch the bus, the passenger runs at a constant speed v towards the bus, What must be the minimum speed of the passenger so that he may catch the bus?

బస్సుకు 48 మీటర్ల దూరంలో ఒక మనిషి నిలబడి ఉన్నాడు. బస్సు బయలుదేరగానే, ఆ వ్యక్తి 10 మీ/సె స్థిర వేగంతో బస్సు వైపు పరిగెత్తాడు. బస్సు నిశ్చల స్థితి నుండి 1 మీ/సె^2 త్వరణం తో చలిస్తుంది. ఆ వ్యక్తి ఆ బస్సు పట్టుకోగల కనీస సమయాన్ని లెక్కించండి

A man is at a distance from the bus: when the bus begins to move with a constant acceleration 'a', with what minimum velocity the man should run towards the bus so as to catch it?

A man runs at a speed of 4 ms to overtake a standing bus. When he is 6m behind the door (a ) the bus moves forward and continues moving with constant acceleration of 1.2 m/s^2 . The time after which the man is able to catch the bus is :

A man is 10m behind the bus stop when the bus left the stop, with an acceleration of 2m//s^(2) . What should be the minimum uniform speed of man so that he may catch the bus? If he runs with required minimum speed then how much time it will take to catch the bus?

A bus starts moving with `1 ms^(-2)` acceleration. A man 48 m away from bus starts moving with 10 ms to catch the bus, then man will reach to bus after ........

A man is 25 m behind a bus, when bus starts accelerating at 2 ms^(-2) and man starts moving with constant velocity of 10 ms^(-1) . Time taken by him to board the bus is

A bus begins to move with an acceleration of 1 `ms^(-2)`. A man who is 48 m behind the bus starts running at 10 `ms^(-1)` to catch the bus. The man will be able to catch the bus after

AIIMS PREVIOUS YEAR PAPERS-AIIMS 2016-PHYSICS
  1. A man is at a distance of 6 m from a bus. The bus begins to move with ...

    08:39

    |

  2. If vecA and vecB are non-zero vectors which obey the relation |vecA+ve...

    Text Solution

    |

  3. In a Fraunhofer diffraction at single slit of width d with incident li...

    Text Solution

    |

  4. A body of mass 60 kg suspended by means of three strings, P, Q and R a...

    03:42

    |

  5. The angular amplitude of a simple pendulum is theta(0). The maximum te...

    02:47

    |

  6. Three identical charges are placed at the vertices of an equilateral t...

    Text Solution

    |

  7. A voltmeter of resistance 20000 Omega reads 5 volt. To make it read 20...

    Text Solution

    |

  8. Light wave enters from medium 1 to medium 2. Its velocity in 2^(nd) m...

    02:44

    |

  9. The temperature of a body is increased from −73^(@)C" to "327^(@)C. Th...

    Text Solution

    |

  10. Time period of pendulum, on a satellite orbiting the earth, is

    01:19

    |

  11. Tend identical cells each of potential E and internal resistance r are...

    01:41

    |

  12. Two charge spheres separated at a distance d exert a force F on each o...

    04:56

    |

  13. A gun of mass 10kg fires 4 bullets per second. The mass of each bullet...

    03:34

    |

  14. Water is filled in a container upto height 3m. A small hole of area 'a...

    09:34

    |

  15. A transparent cube of 15 cm edge contains a small air bubble. Its appa...

    04:01

    |

  16. A stone of mass 0.3 kg attched to a 1.5 m long stirng is whirled aroun...

    00:55

    |

  17. A ball is dropped from the top of a building 100 m high. At the same i...

    04:16

    |

  18. If linear momentum if increased by 50% then kinetic energy will be inc...

    02:44

    |

  19. The additional kinetic energy to be provided to a satellite of mass m ...

    02:51

    |

  20. A sphere of mass 10 kg and radius 0.5 m rotates about a tangent. The m...

    Text Solution

    |

Home
Profile
|