Using velocity time graph derive equation …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Mr. X 7 years, 1 month ago
- 1 answers
Test
Related Questions
Posted by Anmol Kumar 1 year, 8 months ago
- 0 answers
Posted by Himanshi Sharma 1 year, 8 months ago
- 0 answers
Posted by Op Garg 1 year, 8 months ago
- 2 answers
Posted by Prasanna Mendon 8 months, 2 weeks ago
- 1 answers
Posted by Moksh Bhatia 1 year, 8 months ago
- 0 answers
Posted by Yash Dwivedi 1 year, 8 months ago
- 3 answers
Posted by Huda Fatima 1 year, 8 months ago
- 2 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Meghna Thapar 6 years, 1 month ago
Let us consider that the object has travelled a distance s in time t under uniform acceleration a. The distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB. Thus, the distance s travelled by the object is given by
s = area OABC (which is a trapezium)
= area of the rectangle OADC + area of the triangle ABDM.
= OA × OC + 1/2 (AD × BD)
Substituting OA = u, OC = AD = t and BD = at, we get
s = u × t + 1/2 (t×at)
or
0Thank You