Advertisements
Advertisements
Question
Figure given below shows a velocity-time graph for a car starting from rest. The graph has three parts AB, BC and CD.
Compare the distance travelled in part BC with the distance travelled in part AB.
Advertisements
Solution
Distance travelled in part BC = Area of the rectangle tBC2t = base × height.
= (2t - t) × vo
= vot
Distance travelled in part AB = Area of the triangle ABt
= (1/2) × base × height
= (1/2) × t × vo
= (1/2) vo t
Therefore, distance travelled in part BC:distance travelled in part AB :: 2:1.
APPEARS IN
RELATED QUESTIONS
Can you suggest the kind of motion of a body from the following distance – time graph?
From the displacement – time graph shown given below calculate:
- Average velocity in the first three seconds.
- Displacement from the initial position at the end of 13 s.
- Time after which the body is at the initial position,
- Average velocity after 8 s.
Ashok starts from his house and travels 3 km to reach his school. On finding the school gates closed, he returns to his house. What is?
The displacement of Ashok
Give the name of the physical quantity that corresponds to the rate of change of displacement.
Assertion: Displacement of a body may be zero when distance travelled by it is not zero.
Reason: The displacement is the shortest distance between initial and final position.
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?
Write a note on the following:
Nautical mile
Obtain a relation for the distance travelled by an object moving with uniform acceleration in the interval between 4th and 5th seconds.
A scalar quantity has ______.
A body is projected up with an initial velocity of u m/s. It goes up to a height, ‘h’ meters in seconds' time. Then it comes back at the point of projection.
Considering negligible air resistance, which of the following statement is true?
