Question

Starting from a stationary position, Rahul paddles his bicycle attain a

velocity of 62 in 30 sec. Find acceleration of bicycle.​

Answer 1

Answer:

Now in the first case, Rahul paddles from the starting point so his initial velocity is zero and attains a velocity of 6 m/s in 30 sec. Therefore his final velocity is 6 m/s and he takes time 30 sec. ⇒a=630=0.2ms−2. So in the first case his acceleration is 0.2ms−2

Answer 2

Question :-

Starting from a Stationary Position, Rahul paddles his Bicycle to attain a Velocity of 6 m/s in 30 Seconds.

Find the Acceleration of the Bicycle.

Solution :-

Given that,

Initial Velocity, u = 0 m/s [As the Bicycle starts from a Stationary Position]

Final Velocity, v = 6 m/s and,

Time taken, t = 30 s

Let the Acceleration, a of the Object be x

$\begin {gathered} \end {gathered}$

Using the First Equation of Motion (for Velocity - Time Relation), we have :

$\begin {aligned} \mathtt {v} &= \mathtt {u + at}\\\\6\ ms^{-1} &= 0\ ms^{-1} + x \times 30\ s \\\\6\ ms^{-1} &= 30x\\\\\therefore\ x &= \dfrac{6}{30}\ ms^{-2} = \bf 0.2\ ms^{-2} \end {aligned}$

The Acceleration, a of the Bicycle is $\bf 0.2\ ms^{-2}$ . . .

$\begin {gathered} \end {gathered}$

Additional Information :-

When an object moves along a Straight Line with Uniform Acceleration, it is possible to Relate its Velocity, Acceleration during Motion, and the distance covered by it in a Time Interval by a Set of Equations known as the Equations of Motion.

$\begin {gathered} \end {gathered}$

They are as Follows :

i. $\mathtt{v = u + at}$

ii. $\mathtt {s = ut + \dfrac{1}{2}\ at^2}$

iii. $\mathtt {2as = v^2 - u^2}$

Where u is the Initial Velocity of the Object which moves with Uniform Acceleration a for Time t

v is the Final Velocity, and

s is the Distance Traveled by the Object