Question

starts from a stationary Rahul
paddle he byclace to
attain
velocity
6 m|s in 30 s
he applies
brakes
such that the velocity of the bicycle comes
down to 4 m|s in the next
5 s . calculate
acceleration of the bicycle
in the both
cases???
​

Answer 1

Answer:

• First case = 0.2 m/s²
• Second case = - 0.4 m/s²

Explanation:

Given :

• Initial velocity of Rahul = 0 m/s
• Time taken = 30 seconds
• Final velocity = 6 m/s
• Time taken for deceleration = 5 seconds
• Final velocity of second case = 4 m/s

To find:

• Acceleration in both the cases

Acceleration in 1st case:

• (Final velocity - Initial velocity) / time

Acceleration = (6-0)/30

Acceleration = 1/5 m/s²

Acceleration = 0.2 m/s²

Acceleration in 2nd case:

Final velocity = 4 m/s

Initial velocity = 6 m/s

Time = 5 seconds

Acceleration = (4-6)/5

Acceleration = - 2/5

Acceleration = - 0.4 m/s²

The acceleration in first case is 0.2 m/s² and acceleration in second case is - 0.4 m/s²

Answer 2

$\huge\sf\red{Answer:}$

According to the question:

⇒ We have to find the Acceleration twice for two parts, firstly let's calculate the acceleration lf the bicycle on both case's that's according to the first case starts from a stationary Rahul paddles he bicycle to attain velocity of 6 m/s in 30 sec he applies brakes and secondly, velocity of the bicycle comes down to 4 m/s in 5 sec. According to both the cases let us assume 'A' as acceleration.

According to Case(1) :

Using formula:

★ ${\sf{\underline{\boxed{\green{\sf{ A = \dfrac{Final \: velocity - Initial \: velocity}{time} }}}}}}$

Calculations:

⇒ $\sf A = \dfrac{6-0}{30}$

⇒ $\sf A = \dfrac{1}{5 \: m/s^2}$

⇒ ${\sf{\underline{\boxed{\green{\sf{ 0.2 \: m/s^2 }}}}}}$

Therefore, 0.2 m/s² is the acceleration according to first case.

According to Case(2) :

Using formula:

★ ${\sf{\underline{\boxed{\green{\sf{A =\dfrac{Final \: velocity - Initial \: velocity}{Time} }}}}}}$

Calculations:

⇒ $\sf A = \dfrac{4-6}{5}$

⇒ $\sf A = \dfrac{- 2}{5}$

⇒ ${\sf{\underline{\boxed{\green{\sf{ - 0.4 m/s^2 }}}}}}$

Therefore, -0.4 m/s² is the acceleration according to second case.