Motion in One Dimension: Problems and Solutions

Motion in One Dimension

Introduction

Motion in one dimension is the study of objects moving along a straight line. This topic introduces fundamental concepts such as displacement, velocity, speed, and acceleration, and applies them to solve real-world and theoretical problems using algebraic and graphical methods.

Key Concepts and Definitions

• Displacement (Δx): The change in position of an object, defined as the final position minus the initial position. Displacement is a vector quantity.

• Distance: The total length of the path traveled by an object, regardless of direction. Distance is a scalar quantity.

• Average Velocity (\(\bar{v}\)): The displacement divided by the time interval during which the displacement occurs. \[ \bar{v} = \frac{\Delta x}{\Delta t} \]

• Average Speed: The total distance traveled divided by the total time taken. \[ \text{Average speed} = \frac{\text{distance}}{\text{time}} \]

• Acceleration (a): The rate of change of velocity with respect to time. \[ a = \frac{\Delta v}{\Delta t} \]

Worked Examples and Applications

Calculating Distance Traveled During Inattention

• Problem: If you are driving at 95 km/h and look away for 2.0 s, how far do you travel?

• Solution: Convert speed to m/s and multiply by time. \[ \Delta x = \bar{v} \Delta t = (95\,\text{km/h}) \left(\frac{1000\,\text{m}}{1\,\text{km}}\right) \left(\frac{1\,\text{h}}{3600\,\text{s}}\right) (2.0\,\text{s}) = 53\,\text{m} \]

Average Speed Calculation

• Problem: What must your car's average speed be to travel 235 km in 2.75 h?

• Solution: \[ \bar{v} = \frac{235\,\text{km}}{2.75\,\text{h}} = 85.5\,\text{km/h} \]

Average Velocity vs. Average Speed

• Problem: A particle moves from 4.8 cm at \(t_1 = -2.0\) s to 8.5 cm at \(t_2 = 4.5\) s. What is its average velocity? Can you calculate its average speed?

• Solution: \[ \bar{v} = \frac{8.5\,\text{cm} - 4.8\,\text{cm}}{4.5\,\text{s} - (-2.0\,\text{s})} = \frac{3.7\,\text{cm}}{6.5\,\text{s}} = 0.57\,\text{cm/s} \] Note: Average speed cannot be calculated without knowing the actual path length.

Negative Velocity and Direction

• Problem: A rolling ball moves from 8.4 cm to -4.2 cm from \(t_1 = 3.0\) s to \(t_2 = 6.1\) s. What is its average velocity?

• Solution: \[ \bar{v} = \frac{-4.2\,\text{cm} - 8.4\,\text{cm}}{6.1\,\text{s} - 3.0\,\text{s}} = \frac{-12.6\,\text{cm}}{3.1\,\text{s}} = -4.1\,\text{cm/s} \] Interpretation: The negative sign indicates motion in the negative direction.

Average Speed vs. Average Velocity in a Round Trip

• Problem: A horse trots 38 m away in 9.0 s, then gallops halfway back in 1.8 s. Calculate (a) average speed and (b) average velocity.

• Solution:

  • Total distance: \(38\,\text{m} + 19\,\text{m} = 57\,\text{m}\)

  • Total displacement: \(38\,\text{m} - 19\,\text{m} = 19\,\text{m}\)

  • Total time: \(9.0\,\text{s} + 1.8\,\text{s} = 10.8\,\text{s}\)

  • \(\text{Average speed} = \frac{57\,\text{m}}{10.8\,\text{s}} = 5.3\,\text{m/s}\)

  • \(\text{Average velocity} = \frac{19\,\text{m}}{10.8\,\text{s}} = 1.8\,\text{m/s}\)

Earth's Average Speed in Orbit

• Problem: The Earth travels about \(10^9\) km in one year. What is its average speed in km/h?

• Solution: \[ \bar{v} = \frac{1 \times 10^9\,\text{km}}{1\,\text{yr}} \left(\frac{1\,\text{yr}}{365.25\,\text{d}}\right) \left(\frac{1\,\text{d}}{24\,\text{h}}\right) = 1.14 \times 10^5\,\text{km/h} \]

Round Trip: Average Speed and Velocity

• Problem: Calculate the average speed and velocity for a round trip: 250 km out at 95 km/h, 1 h lunch, 250 km back at 55 km/h.

• Solution:

  • Total distance: 500 km

  • Total time: \(\frac{250}{95} + 1 + \frac{250}{55} = 2.63 + 1 + 4.55 = 8.18\) h

  • \(\text{Average speed} = \frac{500\,\text{km}}{8.18\,\text{h}} = 61\,\text{km/h}\)

  • \(\text{Average velocity} = 0\) (since displacement is zero)

Acceleration Calculations

• Average Acceleration: \[ a = \frac{\Delta v}{\Delta t} \]

• Example 1: A sports car accelerates from rest to 95 km/h in 4.3 s. \[ a = \frac{95\,\text{km/h} - 0}{4.3\,\text{s}} \left(\frac{1000\,\text{m}}{1\,\text{km}}\right) \left(\frac{1\,\text{h}}{3600\,\text{s}}\right) = 6.1\,\text{m/s}^2 \]

• Example 2: A sprinter accelerates from rest to 9.00 m/s in 1.38 s. \[ a = \frac{9.00\,\text{m/s}}{1.38\,\text{s}} = 6.52\,\text{m/s}^2 \] \[ a = 8.45 \times 10^4\,\text{km/h}^2 \] (converted units)

Constant Acceleration and Kinematic Equations

• Key Equations:

  • \( v = v_0 + at \)

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

  • \( v^2 = v_0^2 + 2a(x - x_0) \)

• Example: A car slows from 28 m/s to rest in 88 m. Find acceleration. \[ a = \frac{0^2 - (28\,\text{m/s})^2}{2 \times 88\,\text{m}} = -4.5\,\text{m/s}^2 \]

Free Fall and Vertical Motion

• Example: A stone is dropped from a cliff and hits the ground after 3.55 s. How high is the cliff? \[ y = y_0 + v_0 t + \frac{1}{2} a t^2 \] \[ y = 0 + 0 + \frac{1}{2} (9.80\,\text{m/s}^2)(3.55\,\text{s})^2 = 61.8\,\text{m} \]

• Example: King Kong falls from 380 m. Find time to fall and velocity before landing. \[ t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 380}{9.80}} = 8.8\,\text{s} \] \[ v = gt = (9.80)(8.8) = 86\,\text{m/s} \]

Projectile Motion (Vertical Only)

• Example: A baseball is hit straight up at 25 m/s. Estimate (a) maximum height, (b) time in air, (c) factors affecting estimate.

  • (a) \( y = \frac{v_0^2}{2g} = \frac{(25)^2}{2 \times 9.8} = 32\,\text{m} \)

  • (b) \( t = \frac{2v_0}{g} = \frac{2 \times 25}{9.8} = 5.1\,\text{s} \)

  • (c) Estimate factors: Air resistance ignored, "almost straight up" not exact, assumes caught at same height.

Graphical Analysis of Motion

• Velocity-Time Graphs: The slope of a velocity-time graph gives acceleration. The area under the curve gives displacement.

• Example: For a train's velocity-time graph:

  • Greatest velocity at the highest point (e.g., t = 48 s).

  • Constant velocity where the graph is flat (e.g., t = 90–108 s).

  • Constant acceleration where the graph is a straight line with nonzero slope.

  • Greatest acceleration where the slope is steepest (e.g., t = 65–83 s).

Summary Table: Key Kinematic Quantities

Quantity	Symbol	Definition/Formula	SI Unit
Displacement	\(\Delta x\)	\(x_2 - x_1\)	m
Distance	-	Sum of path lengths	m
Average Velocity	\(\bar{v}\)	\(\frac{\Delta x}{\Delta t}\)	m/s
Average Speed	-	\(\frac{\text{distance}}{\text{time}}\)	m/s
Acceleration	\(a\)	\(\frac{\Delta v}{\Delta t}\)	m/s2

Additional info: These problems and solutions cover the essential methods for analyzing one-dimensional motion, including both constant and variable acceleration, and interpreting motion graphs. The examples provided are representative of typical introductory physics problems and are foundational for further study in kinematics and dynamics.