Class 9
Chapter 8 Class 9 - Motion

Slide7.JPG

Slide8.JPG
Slide9.JPG

Go Ad-free

Transcript

Example 8.4 Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases. Case 1 He starts from stationary position, and attains speed of 6 m/s in 30s Since he starts from stationary position Initial velocity = u = 0 m/s Final velocity = v = 6 m/s Time taken = t = 30 s Finding Acceleration Acceleration = (πΉπ‘–π‘›π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ βˆ’ πΌπ‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦)/π‘‡π‘–π‘šπ‘’ = (𝑣 βˆ’ 𝑒)/𝑑 = (6 βˆ’ 0)/30 = 6/30 = 0.2 m/s2 Case 2 He applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s Here, Final velocity in Case 1 becomes the initial velocity, Initial velocity = u = 6 m/s Final velocity = v = 4 m/s Time taken = 5 s Finding Acceleration Acceleration = (πΉπ‘–π‘›π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ βˆ’ πΌπ‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦)/π‘‡π‘–π‘šπ‘’ = (𝑣 βˆ’ 𝑒)/𝑑 = (4 βˆ’ 6)/5 = (βˆ’2)/5 = βˆ’ 0.4 m/s2 Therefore, Acceleration in Case 1 = 0.2 m/s2 Retardation in Case 2 = βˆ’0.4 m/s2

Maninder Singh's photo - Co-founder, Teachoo

Made by

Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 14 years and a teacher from the past 18 years. He teaches Science, Economics, Accounting and English at Teachoo