Comprehensive Study Notes on Describing Motion and Kinematics

Accelerated Motion at Constant Speed

• Problem Statement (Q3): A girl is riding her scooter and observes that the speedometer reading remains constant. The question explores whether it is possible for the scooter to be accelerating under these conditions.

• Conceptual Analysis:

  • Speedometer Function: A speedometer measures the magnitude of velocity, which is the instantaneous speed. It does not account for the direction of travel.

  • Velocity as a Vector: Velocity is defined by both a magnitude (speed) and a direction.

  • Acceleration Definition: Acceleration ($a$) is defined as the rate of change of velocity ($v$). Mathematically, this is expressed as:

  • $a = \frac{dv}{dt}$

  • Since velocity is a vector, acceleration occurs if either the magnitude changes, the direction changes, or both change.

• Conclusion: Yes, it is possible. If the girl is riding her scooter in a curved path or a circle while maintaining a constant speed, the direction of her velocity vector is continuously changing. This change in direction constitutes centripetal acceleration, even though the speedometer reading (speed) does not change.

Kinematic Calculations: Car Starting from Rest

• Problem Statement (Q4): A car starts from rest and its velocity reaches $24\,m\,s^{-1}$ in a duration of $6\,s$. The task is to calculate the average acceleration and the total distance traveled during this time.

• Given Data:

  • Initial velocity ($u$): $0\,m\,s^{-1}$

  • Final velocity ($v$): $24\,m\,s^{-1}$

  • Time interval ($t$): $6\,s$

• Calculation of Average Acceleration:

  • Use the first equation of motion: $v = u + at$, or the definition of acceleration: $a = \frac{v - u}{t}$.

  • $a = \frac{24\,m\,s^{-1} - 0\,m\,s^{-1}}{6\,s}$

  • $a = 4\,m\,s^{-2}$

• Calculation of Distance Traveled:

  • Use the second equation of motion: $s = ut + \frac{1}{2}at^2$

  • $s = (0\,m\,s^{-1} \times 6\,s) + \frac{1}{2}(4\,m\,s^{-2})(6\,s)^2$

  • $s = 0 + 2 \times 36\,m$

  • $s = 72\,m$

  • Alternatively, using $v^2 = u^2 + 2as$:

  • $(24\,m\,s^{-1})^2 = (0\,m\,s^{-1})^2 + 2(4\,m\,s^{-2})(s)$

  • $576 = 8s$

  • $s = 72\,m$

Kinematic Calculations: Decelerating Motorbike

• Problem Statement (Q5): A motorbike moving with an initial velocity of $28\,m\,s^{-1}$ and constant acceleration comes to a complete stop after traveling a distance of $98\,m$. The task is to find the acceleration and the time taken to stop.

• Given Data:

  • Initial velocity ($u$): $28\,m\,s^{-1}$

  • Final velocity ($v$): $0\,m\,s^{-1}$ (comes to a stop)

  • Distance ($s$): $98\,m$

• Calculation of Acceleration:

  • Use the third equation of motion: $v^2 = u^2 + 2as$

  • $(0\,m\,s^{-1})^2 = (28\,m\,s^{-1})^2 + 2a(98\,m)$

  • $0 = 784 + 196a$

  • $196a = -784$

  • $a = -4\,m\,s^{-2}$

  • The negative sign indicates deceleration (retardation).

• Calculation of Time Taken:

  • Use the first equation of motion: $v = u + at$

  • $0 = 28\,m\,s^{-1} + (-4\,m\,s^{-2})t$

  • $4t = 28$

  • $t = 7\,s$

Graphical Analysis: Position-Time Graphs

• Problem Statement (Q6): Based on Fig. 4.27, which displays a position-time graph of two objects, A and B, moving along parallel tracks in the same direction. The question asks if objects A and B ever possess equal velocity.

• Principles of Graphical interpretation:

  • The Slope Rule: In a position-time ($y-t$ or $x-t$) graph, the velocity of the object at any instant is equal to the slope of the graph at that point.

  • Constant Velocity Graph: A straight line on a position-time graph indicates a constant slope, meaning constant velocity.

  • Variable Velocity Graph: A curved line indicates a changing slope, meaning the object is accelerating or decelerating.

• Analysis of Fig 4.27:

  • Object A: Represented by a line passing through the origin ($0,0$). If it is a straight line, its velocity is constant.

  • Object B: Starts at a position $y_0$ above the origin.

  • Comparison: To determine if they have equal velocity, one must identify a point in time where the tangent to the curve of object B has the exact same gradient (slope) as the line/curve of object A.

• Justification:8

  • If there is a point where the lines are parallel (slopes are identical), then the velocities are equal at that specific moment.

  • Usually, when a curved path (accelerating object) intersects or follows a straight path (constant velocity object), there is a point where their instantaneous slopes match. If the lines never reach a state of being parallel, their velocities remain different throughout the observed motion.