Kinematic Analysis of Uniformly Accelerating Bicycle Motion

The scenario involves a bike moving along a straight road, which indicates linear motion or rectilinear kinematics where displacement and distance are identical in magnitude provided there is no change in direction.
Initial state: The bike is stated to "start from rest." In physics, this is a standard phrase indicating that the initial velocity ( $u$ ) is exactly zero at the start of the time interval considered.
Motion type: The problem specifies "uniform acceleration." This means the rate of change of velocity ( $a$ ) remains constant throughout the entire motion, allowing for the application of standard kinematic equations.
Parameters Provided in the Transcript:
- Displacement ( $s$ ): $100\,m$
- Final velocity ( $v$ ) achieved after the displacement: $120\,cm\,s^{-1}$
- Starting condition: "From rest," implying $u = 0\,cm\,s^{-1}$

To solve kinematic equations accurately, all physical variables must be expressed in a consistent system of units (either SI units or CGS units).
System 1: Converting displacement to centimeters (centimeter-gram-second units):
- Given $s = 100\,m$
- Conversion factor: $1\,m = 100\,cm$
- Calculation: $100\,m \times 100\,cm/m = 10000\,cm$
System 2: Converting velocity to meters (meter-kilogram-second units):
- Given $v = 120\,cm\,s^{-1}$
- Conversion factor: $100\,cm = 1\,m$
- Calculation: $\frac{120}{100} = 1.2\,m\,s^{-1}$
Selection: For this detailed guide, calculations will be performed primarily using centimeters to maintain direct consistency with the velocity value provided in the prompt ( $120\,cm\,s^{-1}$ ).

The following equations describe the motion of an object experiencing constant acceleration:
1. $v = u + at$ , relating velocity and time.
2. $s = ut + \frac{1}{2}at^2$ , relating displacement and time.
3. $v^2 = u^2 + 2as$ , relating velocity and displacement independent of time.
4. $s = \frac{(u + v)}{2} \times t$ , calculating displacement using average velocity.
Where:
- $u$ = Initial velocity
- $v$ = Final velocity
- $a$ = Acceleration
- $s$ = Displacement/Distance
- $t$ = Time elapsed

Objective: Determine the bike's uniform acceleration ( $a$ ).
Known Data:
- $u = 0\,cm\,s^{-1}$
- $v = 120\,cm\,s^{-1}$
- $s = 10000\,cm$
Choice of Equation: The third kinematic equation $v^2 = u^2 + 2as$ is chosen because it relates velocity and distance without requiring time ( $t$ ).
Algebraic Rearrangement:
- $a = \frac{v^2 - u^2}{2s}$
Substitution and Calculation:
- $a = \frac{(120\,cm\,s^{-1})^2 - (0\,cm\,s^{-1})^2}{2 \times 10000\,cm}$
- $a = \frac{14400\,cm^2\,s^{-2}}{20000\,cm}$
- $a = 0.72\,cm\,s^{-2}$
Summary: The bike's constant acceleration is $0.72\,cm\,s^{-2}$ .

Objective: Determine the total time ( $t$ ) the bike took to cover $100\,m$ .
Known Data:
- $u = 0\,cm\,s^{-1}$
- $v = 120\,cm\,s^{-1}$
- $a = 0.72\,cm\,s^{-2}$
Choice of Equation: The first kinematic equation $v = u + at$ is the most direct method to solve for $t$ .
Algebraic Rearrangement:
- $t = \frac{v - u}{a}$
Substitution and Calculation:
- $t = \frac{120\,cm\,s^{-1} - 0\,cm\,s^{-1}}{0.72\,cm\,s^{-2}}$
- $t = \frac{120}{0.72}\,s$
- $t = 166.67\,s$
Alternative Verification (Using Displacement):
- Using $s = \frac{(u + v)}{2} \times t$
- $10000 = \frac{(0 + 120)}{2} \times t$
- $10000 = 60 \times t$
- $t = \frac{10000}{60} = 166.67\,s$
Summary: The bike takes approximately $166.67\,s$ to reach the specified velocity over the stated distance.

Objective: Find the instantaneous velocity of the bike precisely $3\,s$ after it starts from rest.
Condition: The acceleration ( $a$ ) calculated in Part (a) remains constant throughout the motion.
Known Data for this step:
- $u = 0\,cm\,s^{-1}$
- $a = 0.72\,cm\,s^{-2}$
- $t = 3\,s$
Choice of Equation: $v = u + at$
Substitution and Calculation:
- $v = 0\,cm\,s^{-1} + (0.72\,cm\,s^{-2} \times 3\,s)$
- $v = 2.16\,cm\,s^{-1}$
Summary: At the three-second mark, the bike is traveling at a velocity of $2.16\,cm\,s^{-1}$ .