Kinematics Exam Practice – Study Notes and Worked Examples

Kinematics: Motion in One Dimension

Introduction

Kinematics is the branch of physics that describes the motion of objects without considering the causes of motion. It involves concepts such as displacement, velocity, acceleration, and time. The following study notes cover key definitions, formulas, and example problems relevant to introductory college physics.

Key Concepts in Kinematics

Displacement, Distance, and Position

• Displacement: The change in position of an object; a vector quantity with both magnitude and direction.

• Distance: The total length of the path traveled by an object; a scalar quantity (no direction).

• Position: The location of an object at a particular time, usually given relative to an origin.

Formula for Displacement:

$\Delta x = x_f - x_i$

• $x_f$ = final position

• $x_i$ = initial position

Velocity and Speed

• Velocity: The rate of change of displacement; a vector quantity.

• Speed: The rate of change of distance; a scalar quantity.

Average Velocity:

$\bar{v} = \frac{\Delta x}{\Delta t}$

Average Speed:

$\text{Average speed} = \frac{\text{Total distance}}{\Delta t}$

Acceleration

• Acceleration: The rate of change of velocity; a vector quantity.

Average Acceleration:

$\bar{a} = \frac{\Delta v}{\Delta t}$

Equations of Motion (Constant Acceleration)

• $v = v_0 + at$

• $x = x_0 + v_0 t + \frac{1}{2} a t^2$

• $v^2 = v_0^2 + 2a(x - x_0)$

Where:

• $v$ = final velocity

• $v_0$ = initial velocity

• $a$ = acceleration

• $t$ = time

• $x$ = final position

• $x_0$ = initial position

Worked Example Problems

1. Relative Motion and Reference Frames

Scenario: Two people observe a toy car moving along the floor. Tammy says, "the velocity of the car is +5.00 cm/s." Timmy says, "No, the velocity of the car is -5.00 cm/s." Both are correct if they are using different reference frames (e.g., facing opposite directions).

• Key Point: Velocity is relative to the observer's frame of reference.

• Application: Always specify the direction and reference frame when describing velocity.

2. Describing Motion: Displacement, Path Length, Speed, and Velocity

Scenario: A bug crawls from 3.61 cm to 4.17 cm in 12.35 seconds.

• Displacement: $4.17\ \text{cm} - 3.61\ \text{cm} = 0.56\ \text{cm}$

• Path Length: $0.56\ \text{cm}$ (if the path is straight)

• Average Speed:$\frac{0.56\ \text{cm}}{12.35\ \text{s}} = 0.045\ \text{cm/s}$

• Average Velocity: $0.045\ \text{cm/s}$ (if motion is in a straight line)

3. Multi-Segment Motion: Calculating Total Displacement and Average Speed

Scenario: A person runs 62.7 meters east, then 43.9 meters west in 20.5 seconds.

• Total Displacement: $62.7\ \text{m} - 43.9\ \text{m} = 18.8\ \text{m}$ east

• Total Distance:$62.7\ \text{m} + 43.9\ \text{m} = 106.6\ \text{m}$

• Average Speed:$\frac{106.6\ \text{m}}{20.5\ \text{s}} = 5.20\ \text{m/s}$

• Average Velocity: $\frac{18.8\ \text{m}}{20.5\ \text{s}} = 0.917\ \text{m/s}$ east

4. Interpreting Position-Time Graphs

Scenario: Given a function $x = 6.12\ \text{m/hr}^2 \cdot t^2 + 13.6\ \text{m}$, sketch the graph and describe the motion.

• Key Point: The equation is quadratic in time, indicating constant acceleration.

• Initial Position:$x_0 = 13.6\ \text{m}$

• Acceleration: $a = 2 \times 6.12 = 12.24\ \text{m/hr}^2$

• Application: The object starts at 13.6 m and accelerates as time increases.

5. Changing Time Intervals: Effects on Final Position and Velocity

Scenario: An object travels for 7.0 hours; what happens to final position and velocity if it travels for only 3.5 hours?

• Key Point: For constant acceleration, position depends on $t^2$ and velocity on $t$.

• Application: Halving the time reduces final velocity by half and final position by a factor of four (since $x \propto t^2$).

6. Motion with Piecewise Functions

Scenario: An object moves as $x = (1.2\ \text{m/s}^2)t^2 + (2.4\ \text{m/s})t$.

• Key Point: This is a uniformly accelerated motion with initial velocity $2.4\ \text{m/s}$ and acceleration $2.4\ \text{m/s}^2$.

• Graphing: The $x$ vs. $t$ graph is a parabola; the $v$ vs. $t$ graph is a straight line with slope $2.4\ \text{m/s}^2$.

7. Free Fall and Vertical Motion

Scenario: A ball is dropped from a 12.6 m tall building. How long does it take to hit the ground?

• Key Point: Use $y = y_0 + v_0 t + \frac{1}{2} a t^2$ with $v_0 = 0$, $a = -9.8\ \text{m/s}^2$.

• Solution: $0 = 12.6 - 4.9 t^2 \implies t = \sqrt{\frac{12.6}{4.9}} \approx 1.6\ \text{s}$

8. Bouncing Ball: Rebound Height

Scenario: A soccer ball is kicked straight up and returns to the ground. If the initial velocity is $10.2\ \text{m/s}$, how much time does it take to return?

• Key Point: Time to rise equals time to fall; total time $t = \frac{2v_0}{g}$

• Solution: $t = \frac{2 \times 10.2}{9.8} \approx 2.08\ \text{s}$

9. Application: Cartoon Physics

Scenario: Wile E. Coyote drops a large ball from a 17.3 m tall cliff. How long does it take to hit the ground?

• Key Point: Use free fall equation: $y = y_0 + v_0 t + \frac{1}{2} a t^2$.

• Solution: $0 = 17.3 - 4.9 t^2 \implies t = \sqrt{\frac{17.3}{4.9}} \approx 1.88\ \text{s}$

Summary Table: Kinematic Quantities

Quantity	Symbol	Definition	SI Unit
Displacement	$\Delta x$	Change in position	meter (m)
Velocity	$v$	Rate of change of displacement	meter/second (m/s)
Acceleration	$a$	Rate of change of velocity	meter/second2 (m/s2)
Time	$t$	Duration of motion	second (s)

Additional info:

• All problems assume motion in a straight line (one dimension).

• Air resistance is neglected in free fall problems.

• For more complex motions, break the motion into segments and analyze each separately.