Question

Starting from a stationary position, Rahul Paddles his bicycle to attain a Velocity of 6 m s $^-1$ in 39s. 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 cases.

Answer 1

$\tt \: First \: case :$

Given,

Initial Velocity(u) =$\tt 0 \: m {s}^{ - 1}$

Final Velocity (v) = $\tt 6 \: m {s}^{ - 1}$

Time = 39 s

Using,

$\huge \boxed { \tt a = \frac{v - u}{t} }$

$\tt \longrightarrow a = \frac{6 \: m \: {s}^{ - 1 } - 0 \: m \: {s}^{ - 1} }{39 \: s} \\ \\ \tt \longrightarrow a = \frac{6 \: m \: {s}^{ - 1} }{39 \: s} \\ \\ \tt \longrightarrow a = 0.15 \: m \: {s}^{ - 2}$

$\tt Second \: case :$

Given,

Initial Velocity (u) = $\tt 6 \: m {s}^{ - 1}$

Final Velocity (v) = $\tt 4 \: m {s}^{ - 1}$

Time = 5 s

$\huge \boxed { \tt a = \frac{v - u}{t} }$

$\tt \longrightarrow a = \frac{4 \: m \: {s}^{ - 1 } - 6 \: m \: {s}^{ - 1} }{5 \: s} \\ \\ \tt \longrightarrow a = \frac{ - 2 \: m \: {s}^{ - 1} }{5\: s} \\ \\ \tt \longrightarrow a = - 0.4 \: m \: {s}^{ - 2}$

The acceleration of bicycle in first case = $\bf 0.15 m \: s^{-2}$ and in the second case = $\bf -0.4 m \: s^{-2}$

Additional Information

• Rate of Change of Distance is known as Speed.

• The speed of body which is applied in beginning is known as Initial Speed and is denoted by u.

• The speed of body which is acquired after the body starts Moving is know as Final Speed and is represented by v.

• Rate of change of Velocity is known as Acceleration.

• Negative Acceleration is known as Retardation.

Important Formula

$\tt Speed = \frac{Distance}{Time}$

$\tt Average Speed = \frac{Total \: distance}{Total \: time}$

$\tt v = u + at$

$\tt v^{2} - u^{2} = 2as$

$\tt S = ut + 1/2at^{2}$

Answer 2

Starting from initial position, Rahul paddles his bicycle to attain a velocity of 6 m/s in 39 seconds.

As, the Rahul starts from rest mean his initial velocity is 0 m/s, his final velocity is 6 m/s and time taken by him is 39 seconds.

We have to find the acceleration of the bicycle.

Using the First Equation Of Motion:

v = u + at

Substitute the known values in the above formula,

→ 6 = 0 + a(39)

→ 6 = 39a

→ 0.15 = a

Therefore, the acceleration of the bicycle is 0.15 m/s².

In second part of the question it is given that, Rahul applied brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5 sec.

Now, his final velocity in first case is equal to his initial velocity i.e. 6 m/s, final velocity is 4 m/s and time taken is 5 seconds.

Again we have to find the acceleration of the bicycle.

Using the First Equation Of Motion:

v = u + at

Substitute the values,

→ 4 = 6 + a(5)

→ 4 - 6 = 5a

→ -2 = 5a

→ -0.4 = a

(Negative sign shows retardation)

Therefore, the acceleration of the bicycle is 0.4 m/s².