Question

Starting from a stationary position. Rahul paddles his bicycle to attain a velocity of 6ms-1 in 30s.Then he applies breaks such that the velocity of the bicycle comes down to 4ms-1 in the next 5s. Calculate the acceleration of the bicycle in both the case.

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Answer 1

Answer:

$For first case:</p><p>Initial velocity =0</p><p>Final velocity =60m/s</p><p>Time =30sec</p><p>v=u+at</p><p>6=0+a×30</p><p>6=30a</p><p>a=306=0.2</p><p>a=0.2m/s2</p><p>For second case:</p><p>Initial velocity =6m/s</p><p>Final velocity =4m/s</p><p>Time taken =5sec</p><p>v=u+at</p><p>4=6+a×5</p><p>4−6=5a</p><p>−2=5a</p><p>a=−52</p><p>a=−0.4m/s2</p><p></p><p>$

Explanation:

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Answer 2

Explanation:

Case I

Initial velocity u = 0 m/s

Final velocity v =6 m/s

Time taken = 30 s

Acceleration = v-u/t

= (6 m/s - 0 m/s/30 s)

= 6 m/s/30 s

= (6/30) m/s²

= 1/5 m/s²

= 0.2 m/s²

Case II

Initial velocity u= 6 m/s

Final velocity v = 4 m/s

Time taken = 5 s

Acceleration = v-u/t

= 4 m/s - 6 m/s/5 s

= -2 m/s/5s

= (-2/5) m/s²