Starting from a stationary position, Rahul …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Abhsiehk Kumar 6 years, 6 months ago
- 3 answers
Test
Related Questions
Posted by Himanshi Sharma 1 year, 4 months ago
- 0 answers
Posted by Prasanna Mendon 5 months ago
- 1 answers
Posted by Yash Dwivedi 1 year, 4 months ago
- 3 answers
Posted by Op Garg 1 year, 4 months ago
- 2 answers
Posted by Moksh Bhatia 1 year, 5 months ago
- 0 answers
Posted by Anmol Kumar 1 year, 4 months ago
- 0 answers
Posted by Huda Fatima 1 year, 4 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
Sia ? 6 years, 6 months ago
Case I: In this case Initial velocity of Rahul is equal to zero i.e, u = 0
Final velocity, v = 6 m/s and Time taken, t = 30 s
We know that, Acceleration = {tex}\frac { \text { Change in velocity } } { \text { Time taken } } {/tex}
{tex}= \frac { \text { Final velocity } - \text { Initial velocity } } { \text { Time taken } }{/tex}
{tex}= \frac { v - u } { t } = \frac { 6 - 0 } { 30 }{/tex} = 0.2 m/s2
Case II: In this case Initial velocity, u = 6 m/s
Final Velocity = {tex}\frac { v - u } { t } = \frac { 4 - 6 } { 5 }{/tex}
= -0.4 m/s
Hence, the acceleration in both the cases are 0.2 m/s2 and -0.4 m/s2 respectively.
0Thank You