Kinematics 1 G Graphs Of Velocity Answers

Let's unravel the world of kinematics, focusing specifically on analyzing motion under constant acceleration due to gravity using velocity-time graphs. Kinematics, the branch of physics that describes the motion of objects without considering the forces that cause that motion, provides a powerful framework for understanding how things move. Velocity-time graphs are essential tools in this context, offering a visual representation of an object's velocity as it changes over time. Understanding these graphs, especially when dealing with constant acceleration scenarios like objects in free fall near the Earth's surface (where the acceleration is approximately g = 9.8 m/s²), is crucial for solving a wide range of physics problems.

Understanding Velocity-Time Graphs

A velocity-time graph plots the velocity of an object on the y-axis against time on the x-axis. The shape and features of this graph tell us a lot about the object's motion:

• Slope: The slope of a velocity-time graph represents the acceleration of the object. A positive slope indicates acceleration in the positive direction, a negative slope indicates acceleration in the negative direction, and a zero slope indicates constant velocity (no acceleration). The steeper the slope, the greater the magnitude of the acceleration.

• Area Under the Curve: The area under the velocity-time graph represents the displacement of the object. Displacement is the change in position of the object. If the area is above the x-axis, the displacement is positive; if it's below the x-axis, the displacement is negative.

• Y-intercept: The y-intercept of the graph represents the initial velocity of the object at time t=0.

Kinematics Equations for Constant Acceleration

Before diving into examples and graph interpretation, let's review the fundamental kinematic equations that govern motion under constant acceleration:

1. v = u + at (Final velocity = Initial velocity + (Acceleration * Time))
2. s = ut + (1/2)at² (Displacement = (Initial velocity * Time) + (1/2 * Acceleration * Time²))
3. v² = u² + 2as (Final velocity² = Initial velocity² + (2 * Acceleration * Displacement))
4. s = (u+v)t/2 (Displacement = (Initial velocity + Final velocity) * Time / 2)

Where:

• v = final velocity
• u = initial velocity
• a = acceleration (in our case, often g, the acceleration due to gravity)
• t = time
• s = displacement

These equations are your toolkit for solving problems involving constant acceleration. Mastery of these equations, coupled with understanding velocity-time graphs, will allow you to tackle complex kinematic problems with confidence.

Analyzing Motion Under Gravity Using Velocity-Time Graphs

Now, let's apply our knowledge to scenarios where the only acceleration is due to gravity. We will assume that air resistance is negligible for simplicity. A classic example is a projectile thrown vertically upwards.

Case 1: Object Thrown Vertically Upwards

Imagine throwing a ball straight up into the air. Here's how to break down the motion and its representation on a velocity-time graph:

1. Initial Stage (Ascent):

   • The ball starts with an initial upward velocity (u).
   • Gravity acts downwards, causing a negative acceleration (-g).
   • The ball's upward velocity decreases linearly with time. On the graph, this is represented by a straight line with a negative slope.

2. Highest Point:

   • At the highest point, the ball momentarily stops. Its velocity is zero (v = 0).
   • On the graph, this is where the line intersects the x-axis (velocity = 0). This point represents the time at which the ball reaches its maximum height.

3. Final Stage (Descent):

   • The ball now starts to fall downwards, accelerating due to gravity (g).
   • Its velocity increases linearly in the downward direction. We conventionally define downward motion as negative, so the velocity becomes increasingly negative. On the graph, this is represented by a straight line with a negative slope (continuing from the ascent line).

Key features of the Velocity-Time Graph:

• Straight Line: The graph is a straight line because the acceleration is constant (g).
• Negative Slope: The slope is negative, indicating that the acceleration is downwards (opposite to the initial upward velocity).
• X-intercept: The x-intercept represents the time at which the ball reaches its highest point (velocity is zero).
• Area Above and Below the X-axis: The area above the x-axis represents the upward displacement (ascent), and the area below the x-axis represents the downward displacement (descent). Since the ball returns to its starting point, the total displacement is zero, meaning the area above the x-axis is equal to the area below the x-axis.

Example Problem:

A ball is thrown vertically upwards with an initial velocity of 15 m/s. Ignoring air resistance, determine:

• a) The time it takes to reach its highest point.
• b) The maximum height reached.
• c) The velocity of the ball when it returns to the starting point.
• d) Sketch a velocity-time graph representing the motion.

Solution:

a) Time to reach the highest point:

• u = 15 m/s
• v = 0 m/s (at the highest point)
• a = -9.8 m/s²
• Using the equation v = u + at:
  • 0 = 15 + (-9.8) * t
  • t = 15 / 9.8 ≈ 1.53 seconds

b) Maximum Height Reached:

• Using the equation v² = u² + 2as:
  • 0² = 15² + 2 * (-9.8) * s
  • 0 = 225 - 19.6s
  • s = 225 / 19.6 ≈ 11.48 meters

c) Velocity when it returns to the starting point:

• The time for the entire flight (up and down) is twice the time to reach the highest point: 2 * 1.53 = 3.06 seconds.

• Using the equation v = u + at:

  • v = 15 + (-9.8) * 3.06
  • v = 15 - 30 ≈ -15 m/s

• Therefore, the velocity when it returns to the starting point is -15 m/s (same magnitude as the initial velocity but in the opposite direction).

d) Sketching the Velocity-Time Graph:

• The graph will be a straight line with a negative slope.
• The y-intercept will be at 15 m/s (initial velocity).
• The x-intercept will be at 1.53 seconds (time to reach the highest point).
• The graph will continue to t=3.06 seconds where the velocity equals -15 m/s.
• The area above the x-axis (from 0 to 1.53 seconds) will be equal to the area below the x-axis (from 1.53 to 3.06 seconds).

Case 2: Object Dropped from Rest

Now consider an object dropped from rest.

1. Initial Stage:
   • The object starts with zero initial velocity (u = 0).
   • Gravity acts downwards, causing a positive acceleration (g).
   • The object's downward velocity increases linearly with time. On the graph, this is represented by a straight line with a positive slope.

Key features of the Velocity-Time Graph:

• Straight Line: The graph is a straight line because the acceleration is constant (g).
• Positive Slope: The slope is positive, indicating that the acceleration is downwards, in the same direction as the motion we defined as positive.
• Y-intercept: The y-intercept is at 0 m/s (initial velocity).
• Area Under the Curve: The area under the curve represents the displacement (distance fallen).

Example Problem:

An object is dropped from a height of 20 meters. Ignoring air resistance, determine:

• a) The time it takes to reach the ground.
• b) The velocity of the object just before it hits the ground.
• c) Sketch a velocity-time graph representing the motion.

Solution:

a) Time to reach the ground:

• u = 0 m/s
• s = 20 m
• a = 9.8 m/s²
• Using the equation s = ut + (1/2)at²:
  • 20 = (0) * t + (1/2) * (9.8) * t²
  • 20 = 4.9t²
  • t² = 20 / 4.9 ≈ 4.08
  • t = √4.08 ≈ 2.02 seconds

b) Velocity just before hitting the ground:

• Using the equation v = u + at:

  • v = 0 + (9.8) * 2.02
  • v ≈ 19.8 m/s

• Therefore, the velocity just before hitting the ground is approximately 19.8 m/s.

c) Sketching the Velocity-Time Graph:

• The graph will be a straight line with a positive slope.
• The y-intercept will be at 0 m/s (initial velocity).
• At t= 2.02 seconds, the velocity equals 19.8 m/s.
• The area under the line (from 0 to 2.02 seconds) will be equal to 20 meters.

Case 3: Object Thrown Downwards

Consider an object thrown downwards with an initial velocity.

1. Initial Stage:
   • The object starts with some initial downward velocity (u > 0).
   • Gravity acts downwards, causing a positive acceleration (g).
   • The object's downward velocity increases linearly with time. On the graph, this is represented by a straight line with a positive slope.

Key features of the Velocity-Time Graph:

• Straight Line: The graph is a straight line because the acceleration is constant (g).
• Positive Slope: The slope is positive, indicating that the acceleration is downwards, in the same direction as the motion.
• Y-intercept: The y-intercept represents the initial downward velocity.
• Area Under the Curve: The area under the curve represents the displacement (distance fallen).

Example Problem:

An object is thrown downwards from a height of 20 meters with an initial velocity of 5 m/s. Ignoring air resistance, determine:

• a) The time it takes to reach the ground.
• b) The velocity of the object just before it hits the ground.
• c) Sketch a velocity-time graph representing the motion.

Solution:

a) Time to reach the ground:

• u = 5 m/s

• s = 20 m

• a = 9.8 m/s²

• Using the equation s = ut + (1/2)at²:

  • 20 = (5) * t + (1/2) * (9.8) * t²
  • 20 = 5t + 4.9t²
  • 4. 9t² + 5t - 20 = 0

• Using the quadratic formula:

  • t = (-b ± √(b² - 4ac)) / 2a
  • t = (-5 ± √(5² - 4 * 4.9 * -20)) / (2 * 4.9)
  • t = (-5 ± √(25 + 392)) / 9.8
  • t = (-5 ± √417) / 9.8
  • t = (-5 ± 20.42) / 9.8

• We take the positive root:

  • t = (15.42) / 9.8 ≈ 1.57 seconds

b) Velocity just before hitting the ground:

• Using the equation v = u + at:

  • v = 5 + (9.8) * 1.57
  • v ≈ 5 + 15.39 = 20.39 m/s

• Therefore, the velocity just before hitting the ground is approximately 20.39 m/s.

c) Sketching the Velocity-Time Graph:

• The graph will be a straight line with a positive slope.
• The y-intercept will be at 5 m/s (initial velocity).
• At t= 1.57 seconds, the velocity equals 20.39 m/s.
• The area under the line (from 0 to 1.57 seconds) will be equal to 20 meters.

Tips for Interpreting Velocity-Time Graphs

• Pay attention to the sign: The sign of the velocity indicates the direction of motion. Positive is typically "up" or "right," and negative is "down" or "left."
• Understand the slope: The slope represents the acceleration. Constant slope = constant acceleration.
• Calculate the area carefully: Remember to consider the area above and below the x-axis separately when calculating displacement.
• Relate the graph to the physical situation: Visualize the actual motion of the object as you analyze the graph. This will help you catch errors and make sense of the results.

Common Mistakes to Avoid

• Confusing velocity and displacement graphs: Velocity-time graphs show velocity against time, while displacement-time graphs show displacement against time. The interpretation of slope and area is different for each.
• Ignoring the sign: Forgetting to account for the sign of velocity and acceleration can lead to incorrect results, especially when dealing with motion in two directions.
• Incorrectly calculating the area: Be careful when calculating the area under the curve, especially if the graph includes regions above and below the x-axis. You must consider those areas separately.
• Forgetting units: Always include units in your calculations and final answers. This helps you keep track of the quantities and ensure that your results are physically meaningful.

Advanced Concepts and Extensions

The principles discussed here can be extended to more complex scenarios:

• Non-constant Acceleration: While we focused on constant acceleration (g), velocity-time graphs can also be used to analyze motion with non-constant acceleration. In this case, the graph will be curved, and the acceleration at any given time will be the instantaneous slope of the tangent line at that point.
• Two-Dimensional Motion: The concepts can be applied to each component of motion separately in 2D scenarios (e.g., projectile motion with both horizontal and vertical components).
• Calculus Connection: For those familiar with calculus, the velocity is the derivative of the position with respect to time, and the acceleration is the derivative of the velocity with respect to time. The area under the velocity-time curve is the integral of the velocity with respect to time, which gives the displacement.

Conclusion

Understanding velocity-time graphs is fundamental to grasping kinematics, especially motion under the influence of gravity. By mastering the relationship between the graph's slope, area, and intercepts, and by connecting these graphical representations to the kinematic equations, you can confidently solve a wide variety of physics problems. Remember to practice regularly, visualize the physical situations, and pay attention to the details. This will pave the way for a deeper understanding of the fascinating world of motion.