Question

5. Starting from stationary position, dev paddles his bicycle to
attain a velocity of 6 m/s in 30 s. Then he applies brakes
such that the velocity of the bicycle come down to 4 m/s in
the next 5 s. calculates the acceleration of the bicycle in
both cases
(2)​

Answer 1

Starting from initial position, Dev paddles his bicycle to attain a velocity of 6 m/s in 30 seconds.

As, the Dev starts from rest mean his initial velocity is 0 m/s, his final velocity is 6 m/s and time taken by him is 30 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(30)

→ 6 = 30a

→ 0.2 = a

Therefore, the acceleration of the bicycle is 0.2 m/s².

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

Now, his final velocity of 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.

Answer 2

$\huge\tt{Answer:-}$

Case¹ : Dev from stationary position attains a velocity of 6m s^-1 in 30s:-

_________...

$\bf{Given:-}$

• $(t) = 30s$
• $(u) = 0m \ {s}^{-1}$ (Since it was in stationary position)
• $(v) = 6m \ {s}^{-1}$

We know,

Acceleration = (Final Velocity - Initial Velocity)/(Time taken)

Or,

$a = \frac{v-u}{t}$

Putting the values :-

$a = \frac{6m \ {s}^{-1} - 0m \ {s}^{-1}}{30s}$

$\implies a = \frac{6m \ {s}^{-1}}{30s}$

$\implies a = \frac{\cancel{6m} \ {s}^{-1}}{\cancel{30}s}$

$\implies a = \frac{1m }{5} {s}^{-2}$

$\boxed{\implies a = 0.2 m \ {s}^{-2}}$ $...({Ans.}^{1})$

Case² : Applies brakes such that the final velocity of case¹ comes down to 4m s^-1 in 5s:-

$\bf{Given:-}$

• $(t) = 5s$
• $(u) = 6m \ {s}^{-1}$
• $(v) = 4m \ {s}^{-1}$

We know,

We know, Acceleration = (Final Velocity - Initial Velocity)/(Time taken)

Or,

$a = \frac{v-u}{t}$

Putting the values :-

$a = \frac{4m \ {s}^{-1} - 6m \ {s}^{-1}}{5s}$

$\implies a = \frac{-2m \ {s}^{-1}}{5s}$

$\implies a = \frac{-2m}{5} {s}^{-2}$

$\implies a = \frac{\cancel{-2m}}{\cancel{5}} {s}^{-2}$

$\boxed{\implies a = -0. 4m \ {s}^{-2} }$ $...({Ans.}^{2})$