Question

starting from a stationary position. Rahul paddles his bicycle to attain a velocity of 6 m/s in 30s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5s. calculate the acceleration of the bicycle in both the cases.​

Answer 1

Given :

starting from a stationary position. Rahul paddles his bicycle to attain a velocity of 6 m/s in 30s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5s.

To Find :

Calculate the acceleration of the bicycle in both the cases ?

Solution :

In the first case :

• We have :-

• Initial velocity (u) = 0m/s
• Final velocity (v) = 6 m/s
• Time (t) = 30s

We know that,

${ \underline{ \boxed{ \tt{ \purple{ Acceleration = \frac{v - u}{t}}}}}}$

Substituting the given values of u, v and t in the above equation, we get

$\implies{ \rm{a = \frac{6 - 0}{30} }}$

$\implies{ \boxed{ \red{ \rm{a = 0.2 \: m/s ^{2} }}}}$

In Second case :

• We have :-

• Initial velocity (u) = 6m/s
• Final velocity (v) = 4 m/s
• Time (t) = 5s

$\implies{ \rm{Then,a = \frac{4 - 6}{5} }}$

$\implies{ \boxed{ \red{ \rm{a = - 0.4 \: m/s ^{2}}} }}$

The acceleration of the bicycle in the first case is 0.2m/s² and in the second case, it is -0.4m/s².

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Answer 2

GiveN :-

• Starting from a stationary position , Rahul paddles his bicycle to attain a velocity of 6 m/s in 30s.
• Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5s.

To FinD :-

• The acceleration of the bicycle in both the cases.

SolutioN :-

Here initial velocity of Rahul is 0m/s and he attained a velocity of 6m/s in 30s .We need to find the acceleration . We know that the rate of change of velocity is called acceleration .

• C A S E - O N E :-

$\red{\frak{ Given }}\begin{cases} \textsf{ Initial velocity = \textbf{0m/s } .}\\\textsf{ Final velocity =\textbf{ 6m/s }.}\\\textsf{ Time taken =\textbf{30s} .}\end{cases}$

$\red{\bigstar}\underline{\textsf {Using the formula of acceleration :- }}$

$\sf:\dashrightarrow\pink{ Accel^n = \dfrac{v-u}{t}}\\\\\sf\dashrightarrow a = \dfrac{6m/s-0m/s}{30s} \\\\\sf\dashrightarrow a =\dfrac{6}{30} m/s^2 \\\\\sf\dashrightarrow \underset{\blue{\sf Required\ Acceleration}}{\underbrace{\boxed{\pink{\frak{ Acceleration = \dfrac{1}{5} m/s }}}}}$

$\rule{200}2$

• C A S E - T W O :-

$\red{\frak{ Given }}\begin{cases} \textsf{ Initial velocity = \textbf{6m/s } .}\\\textsf{ Final velocity =\textbf{ 4m/s }.}\\\textsf{ Time taken =\textbf{5 s} .}\end{cases}$

$\red{\bigstar}\underline{\textsf {Using the formula of deacceleration :- }}$

$\sf:\dashrightarrow\pink{ Retardl^n = \dfrac{v-u}{t}}\\\\\sf\dashrightarrow a = \dfrac{4m/s-6m/s}{4 s} \\\\\sf\dashrightarrow a =\dfrac{-2}{4} m/s^2 \\\\\sf\dashrightarrow \underset{\blue{\sf Required\ Acceleration}}{\underbrace{\boxed{\pink{\frak{ Retardation = -0.5 m/s }}}}}$