Advertisements
Advertisements
प्रश्न
From the velocity – time graph given below, calculate Average velocity in region CED.
उत्तर
Displacement covered in region ΔCDE = area of ΔCDE
= `1/2xx"ED"xx"CE"`
= `1/2xx(28-16)xx6`
= `1/2xx12xx6`
= 36 m
Average velocity in region CED = `"Total displacement"/"Time"`
= `36/12`
= 3 ms−1
APPEARS IN
संबंधित प्रश्न
What can you say about the motion of a body if:
its displacement-time graph is a straight line ?
Study the speed-time graph of a car given alongside and answer the following questions:
(i) What type of motion is represented by OA ?
(ii) What type of motion is represented by AB ?
(iii) What type of motion is represented by BC ?
(iv) What is the acceleration of car from O to A ?
(v)What is the acceleration of car from A to B ?
(vi) What is the retardation of car from B to C ?
Draw displacement – time graph for the following situation:
When a body is stationary.
Draw velocity – time graph for the following situation:
When a body is moving with variable velocity, but uniform acceleration.
A ball is thrown up vertically and returns back to thrower in 6 s. Assuming there is no air friction, plot a graph between velocity and time. From the graph calculate
- deceleration
- acceleration
- total distance covered by ball
- average velocity.
Given below are the speed -time graphs. Match them with their corresponding motions :
![]() |
(a) Uniformity retared motion |
![]() |
(b) Non-uniformity acceleration |
![]() |
(c) Non-uniform motion |
![]() |
(d) uniform motion |
Sketch the shape of the velocity-time graph for a body moving with:
Uniformly velocity
Draw the distance-time graphs of the bodies P and Q starting from rest, moving with uniform speeds with P moving faster than Q.
State whether true or false. If false, correct the statement.
The velocity – time graph of a particle falling freely under gravity would be a straight line parallel to the x axis.
If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?