Physics Study Guide: Acceleration and Kinematic Motion Graphs

Definition of Acceleration (Q21): Acceleration is formally defined as the change in velocity divided by the time during which the change occurred. It is not the change in distance or displacement alone, nor is it the square of velocity divided by the radius (which describes centripetal acceleration).
The Nature of Acceleration as a Rate (Q22): The rate of change of velocity over time is specifically called acceleration. This distinguishes it from average velocity, distance, or instantaneous velocity.
Relationship Between Acceleration and Velocity (Q23): If an object's acceleration is zero ( $a = 0$ ), its velocity is constant. This means the object is moving at a steady speed in a straight line without speeding up, slowing down, or changing direction.
Magnitude and Units (Q24): If an object increases its speed by $2\,m/s$ every second, this describes a constant acceleration of $2\,m/s^2$ . This scenario characterizes the rate of change rather than the total distance, instantaneous velocity, or total time.

Representing Velocity Changes (Q20): When an object moves at a constant speed and then undergoes acceleration, the motion is represented by specific graphical shapes. According to the transcript answer key, graph (d) correctly represents this sequence of motion.
Calculating Acceleration from Graphs (Q25): The average acceleration of an object is determined by calculating the slope of its velocity-time ( $v-t$ ) graph.
- Example Calculation: Given a velocity-time curve with a peak velocity of $24\,m/s$ over $6\,s$ .
- Formula: $\text{Slope} = \frac{\Delta v}{\Delta t}$
- Calculation: $\frac{24\,m/s}{6\,s} = 4\,m/s^2$ (Option c).

Determining Time from Acceleration (Q26): To find the time ( $t$ ) required to reach a specific velocity given a constant acceleration:
- Initial Condition: Car accelerates from rest ( $v_i = 0\,m/s$ ).
- Constant Acceleration ( $a$ ): $3\,m/s^2$ .
- Final Velocity ( $v_f$ ): $33\,m/s$ .
- Formula: $v_f = v_i + (a \times t)$ , which rearranges to $t = \frac{v_f - v_i}{a}$ .
- Calculation: $t = \frac{33\,m/s - 0\,m/s}{3\,m/s^2} = 11\,s$ (Option b).
Calculating Displacement with Acceleration (Q27): To find the total displacement ( $d$ ) of an object (e.g., a bullet) fired with an initial velocity under constant acceleration:
- Initial Velocity ( $v_i$ ): $10\,m/s$ .
- Time ( $t$ ): $10\,s$ .
- Acceleration ( $a$ ): $5\,m/s^2$ .
- Formula: $d = (v_i \times t) + \frac{1}{2} a \times t^2$
- Calculation: $d = (10\,m/s \times 10\,s) + \frac{1}{2} (5\,m/s^2 \times (10\,s)^2)$
- Calculation: $d = 100\,m + \frac{1}{2} (5 \times 100) = 100\,m + 250\,m = 350\,m$ (Option a).

Impact Velocity in Free Fall (Q28): When an object is dropped (initial vertical velocity is assumed to be $0\,m/s$ unless otherwise stated) and reaches a destination after a specific time:
- Time ( $t$ ): $0.5\,s$ .
- Acceleration due to gravity ( $g$ ): $9.8\,m/s^2$ .
- Mass ( $m$ ): $0.1\,kg$ (Note: mass does not affect the velocity in free fall calculations neglecting air resistance).
- Formula: $v_f = g \times t$
- Calculation: $v_f = 9.8\,m/s^2 \times 0.5\,s = 4.9\,m/s$ (Option d).

Comparing Different Moving Objects (Q29): Acceleration must be calculated for both objects to determine which is greater.
- Object A:
  - Change in velocity: $30\,m/s - 10\,m/s = 20\,m/s$ .
  - Time taken: $4\,s$ .
  - Acceleration ( $a_A$ ): $\frac{20\,m/s}{4\,s} = 5\,m/s^2$ .
- Object B:
  - Change in velocity: $33\,m/s - 22\,m/s = 11\,m/s$ .
  - Time taken: $11\,s$ .
  - Acceleration ( $a_B$ ): $\frac{11\,m/s}{11\,s} = 1\,m/s^2$ .
- Conclusion: The acceleration of Object A is greater (5\,m/s^2 > 1\,m/s^2).