Home
Class 11
PHYSICS
The acceleration of a moving body can be...

The acceleration of a moving body can be found from

A

area under velocity - time graph

B

area under displacement - time graph

C

slope of distance - time graph

D

slope of velocity - time graph

Text Solution

AI Generated Solution

The correct Answer is:
To find the acceleration of a moving body, we can follow these steps: ### Step 1: Understand the Definition of Acceleration Acceleration (a) is defined as the rate of change of velocity (v) with respect to time (t). Mathematically, this is expressed as: \[ a = \frac{dv}{dt} \] ### Step 2: Identify the Relevant Graph In this context, we will use a velocity-time (Vt) graph. The x-axis represents time (t), and the y-axis represents velocity (v). ### Step 3: Relate Acceleration to the Graph In a Vt graph, the acceleration of the moving body can be determined by the slope of the graph. The slope of a curve at any point gives the instantaneous rate of change of velocity with respect to time. ### Step 4: Determine the Slope For a straight line on the Vt graph, the slope can be calculated as: \[ \text{slope} = \frac{\Delta v}{\Delta t} \] This slope is equal to the acceleration (a). ### Step 5: Analyze the Options Now, let's analyze the options given in the question: 1. Area under the Vt graph - This is incorrect because the area under a Vt graph gives displacement, not acceleration. 2. Area under the displacement-time graph - This is also incorrect as it relates to the distance covered, not acceleration. 3. Slope of the distance-time graph - This is incorrect because the slope of a distance-time graph gives velocity, not acceleration. 4. Slope of the velocity-time graph - This is correct as it gives the acceleration. ### Conclusion Thus, the correct answer is that the acceleration of a moving body can be found from the slope of the velocity-time graph. ---

To find the acceleration of a moving body, we can follow these steps: ### Step 1: Understand the Definition of Acceleration Acceleration (a) is defined as the rate of change of velocity (v) with respect to time (t). Mathematically, this is expressed as: \[ a = \frac{dv}{dt} \] ### Step 2: Identify the Relevant Graph In this context, we will use a velocity-time (Vt) graph. The x-axis represents time (t), and the y-axis represents velocity (v). ...
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    DC PANDEY ENGLISH|Exercise (B) Meical entrance special format questions (Mathch the columns)|6 Videos
  • MOTION

    DC PANDEY ENGLISH|Exercise Medical entrances gallery|19 Videos
  • PROJECTILE MOTION

    DC PANDEY ENGLISH|Exercise Level - 2 Subjective|10 Videos

Similar Questions

Explore conceptually related problems

(A) : When external force acting on a body moving on a rough horizontal surface is doubled, the acceleration of the body is doubled. (R ) : The acceleration of a body moving on a rough horizontal surface is proportional to the resultant force acting on it.

Assertion: A negative acceleration of a body can be associated with a 'speeding up' of the body. Reason: Increase in speed of a moving body is independent of its direction of motion.

If acceleration gradient of a moving body along positive x-axis is -4s^(-2) and body starts its motion from rest from x = 5m , having maximum acceleration (magnitude) then velocity of moving body at x = -3m is

STATEMENT - 1 : A positive acceleration of a body can be associated with a slowing down of the body. and STATEMENT - 2 : Acceleration is a vector quantity.

A body is just supported at the face of a cart moving at an acceleration. The acceleration of the cart so that the body does not slide

The direction of angular acceleration of a body moving in a circle in the plane of the paper is .

A body is moving with a velocity 5m//s along a striaght line.A force of 10N is applied to the body perpendicular to the line of motion. If the body is of mass 2kg, then the acceleration of the body along the initial straight line of motion is :

A body starts from rest and moves with uniform acceleration of 5 ms^(-2) for 8 seconds. From that time the acceleration ceases. Find the distance covered in 12s starting from rest.

A body is moving with variable acceleartion (a) along a straight line. The average acceleration of body in time interval t_(1)" to "t_(2) is :-

A : Average velocity can be zero, but average speed of a moving body can not be zero in any finite time interval. R : For a moving body displacement can be zero but distance can never be zero.

DC PANDEY ENGLISH-MOTION IN A PLANE-(C )Medical entrances gallery
  1. A particle moves with constant acceleration along a straight line stre...

    Text Solution

    |

  2. A car covers the first half of the distance between two places at a sp...

    Text Solution

    |

  3. A particle starts moving from rest under uniform acceleration it trave...

    Text Solution

    |

  4. At time t = 0, two bodies A and B at the same point. A moves with cons...

    Text Solution

    |

  5. A motorclist drives from A to B with a uniform speed of 30 kmh^(-1) an...

    Text Solution

    |

  6. A body starts from rest and moves with constant acceleration for t s....

    Text Solution

    |

  7. The acceleration of a moving body can be found from

    Text Solution

    |

  8. A stone falls freely under gravity. It covered distances h1, h2 and h3...

    Text Solution

    |

  9. The motion of a particle in straight line is an example of

    Text Solution

    |

  10. The velocity-time graph of particle comes out to be a non-linear curve...

    Text Solution

    |

  11. A person reaches on a point directly opposite on the other bank of a r...

    Text Solution

    |

  12. A body is thrown vertically upward from a point A 125 m above the grou...

    Text Solution

    |

  13. A particle is moving eastwards with velocity of 5 m//s. In 10 sec the ...

    Text Solution

    |

  14. The velocity-time graph of robber's car and a chasing police car are s...

    Text Solution

    |

  15. Initial speed of an alpha particle inside a tube of length 4m is 1 kms...

    Text Solution

    |

  16. A body X is projected upwards with a velocity of 98 ms^(-1), after 4s,...

    Text Solution

    |

  17. Let r(1)(t)=3t hat(i)+4t^(2)hat(j) and r(2)(t)=4t^(2) hat(i)+3t^()ha...

    Text Solution

    |

  18. The motion of a particle along a straight line is described by equatio...

    Text Solution

    |

  19. A scooter starts from rest have an acceleration of 1 ms^(-2) while a c...

    Text Solution

    |

  20. The displacement x of a particle along a straight line at time t is gi...

    Text Solution

    |