Application Based Questions on Kinematics and Motion Analysis

Kinematic Analysis of a Car in Uniform Straight-Line Motion

• Observational Data of Motion: A car is moving along a straight road, exhibiting change in speed over specific time intervals. The following data points characterize the car's motion:

  • At time $t = 0\,s$, the speed is $4\,m/s$. This represents the initial velocity ($u$).
  • At time $t = 2\,s$, the speed increases to $8\,m/s$.
  • At time $t = 4\,s$, the speed increases to $12\,m/s$.
  • At time $t = 6\,s$, the speed increases to $16\,m/s$.
  • At time $t = 8\,s$, the speed increases to $20\,m/s$.
  • At time $t = 10\,s$, the speed reaches $24\,m/s$. This represents the final velocity ($v$) for the observed interval.

• Analysis of Acceleration:

  • The speed increases by exactly $4\,m/s$ every $2\,s$. This constant rate of change indicates that the car is moving with uniform acceleration.
  • The acceleration ($a$) can be calculated using the formula:

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

• Step-by-Step Calculation of Acceleration:
  • Identify the values from the data set: Final speed $v = 24\,m/s$, Initial speed $u = 4\,m/s$, and Total time $t = 10\,s$.
  • Substitute the values into the formula:

$a = \frac{24 - 4}{10}$

$a = \frac{20}{10}$

$a = 2\,m/s^2$

• Result: The acceleration of the car is $2\,m/s^2$.

Determination of Distance via Speed-Time Relationship

• Distance Travelled in a Specific Interval: To find the distance ($s$) travelled by the car in the first $10\,s$, two primary methods can be employed: the area under the speed-time graph or the kinematic equations of motion.

• Method 1: Integration via Graphical Area: On a speed-time graph, the distance travelled is equal to the area under the curve. For uniform acceleration starting from a non-zero initial speed, this area forms a trapezoid.

  • Formula for the area of a trapezoid:

$\text{Area} = \frac{1}{2} \times (u + v) \times t$

*   Calculation:

$s = \frac{1}{2} \times (4 + 24) \times 10$

$s = \frac{1}{2} \times 28 \times 10$

$s = 14 \times 10$

$s = 140\,m$

• Method 2: Kinematic Equation Application: Using the standard equation for displacement with constant acceleration:

$s = ut + \frac{1}{2}at^2$

*   Calculation:

$s = (4 \times 10) + (\frac{1}{2} \times 2 \times (10)^2)$

$s = 40 + (1 \times 100)$

$s = 140\,m$

• Final Result: The total distance travelled by the car in $10\,s$ is exactly $140\,m$.

Conceptual Principles of Average Velocity and Average Speed

• Distinct Mathematical Definitions:
  • Average Speed: Defined as the total distance travelled divided by the total time taken. Since distance is a scalar quantity and the path length of a moving object is always positive, average speed can never be zero for a body in motion.

$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}$

*   **Average Velocity**: Defined as the net displacement divided by the total time taken. Displacement is a vector quantity representing the straight-line distance between the initial and final positions.

$\vec{v}_{avg} = \frac{\text{Displacement}}{\text{Total Time}}$

• Conditions for Zero Average Velocity:

  • A body can have a zero average velocity if its net displacement is zero. This occurs when the starting point and the ending point of the motion are identical.
  • Crucially, even if the displacement is zero, the total distance (path length) is non-zero, resulting in a non-zero average speed.

• Illustrative Example: Round-Trip or Circular Motion:

  • Scenario: Consider a person who walks from their home to a grocery store located $500\,m$ away and then walks back to their home.
  • Distance calculation: The total distance is $500\,m + 500\,m = 1000\,m$. Since the distance is positive, the average speed is not zero.
  • Displacement calculation: The starting position is home and the final position is home. Therefore, the net displacement is $0\,m$.
  • Conclusion: In this scenario, the average velocity is zero because the numerator (displacement) is zero, while the average speed remains a positive value because the object was in motion and covered a measurable path length.