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प्रश्न
From the displacement – time graph shown given below calculate :
- Velocity between 0 – 2 s.
- Velocity between 8 s – 12 s.
- Average velocity between 5 s – 12 s.
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उत्तर
1. Velocity between 0 – 2 s = `(-(25-10))/2` .......[negative sign because slope is negative]
= `-15/2`
= −7.5 ms−1
2. Velocity between 8 s – 12 s = `(25-20)/(12-8)=5/4`
= 1.25 ms−1
3. Average velocity between 5 s – 12 s
= `"Total displacement from 5 s to 12 s"/"Time"`
= `(25-10)/(12-5)`
= `15/7`
= 2.1 ms−1
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संबंधित प्रश्न
Which of the following is true for displacement?
- It cannot be zero.
- Its magnitude is greater than the distance travelled by the object.
If the displacement of an object is proportional to the square of time, then the object is moving with :
Clarify the difference.
Distance and displacement
Figure shows the velocity-time graph of a particle moving in a straight line.
(i) State the nature of motion of particle.
(ii) Find the displacement of particle at t = 6 s.
(iii) Does the particle change its direction of motion?
(iv) Compare the distance travelled by the particle from 0 to 4 s and from 4 s to 6 s.
(v) Find the acceleration from 0 to 4 s and retardation from 4 s to 6 s.
Calculate the distance and displacement in the following case:
Exercise Problems.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 m and 20 s?
Four cars A, B, C, and D are moving on a levelled road. Their distance versus time graphs are shown in Fig. 8.2. Choose the correct statement.
The displacement is zero when the initial and final positions are ______.
Is displacement a scalar quantity?
A body moves is a circle of radius ‘2R’ what is the distance covered and displacement of the body after 2 complete rounds?
