38. (Il) An object starts from rest and falls under the influence of gravity. Draw graphs of (a) its speed and (b) the
distance it has fallen, as a function of time from / = 0 to
+ = 5.00 s. Ignore air resistance.

To solve this problem, we will use the kinematic equations for uniformly accelerated motion. Since the object is falling under the influence of gravity with no initial velocity and no air resistance, the acceleration will be constant and equal to the acceleration due to gravity (\( g \)), which is approximately \( 9.81 \, \text{m/s}^2 \).

### (a) Graph of Speed as a Function of Time

The speed of the object as a function of time can be calculated using the equation:

\[
v(t) = v_0 + at
\]

Given:
- Initial speed, \( v_0 = 0 \, \text{m/s} \) (the object starts from rest)
- Acceleration, \( a = g = 9.81 \, \text{m/s}^2 \)

Substituting these values into the equation:

\[
v(t) = 0 + (9.81 \, \text{m/s}^2) \cdot t = 9.81t \, \text{m/s}
\]

The speed graph will be a straight line starting from the origin, with a slope equal to \( 9.81 \, \text{m/s}^2 \). At \( t = 5.00 \, \text{s} \), the speed will be:

\[
v(5.00) = 9.81 \times 5.00 = 49.05 \, \text{m/s}
\]

### (b) Graph of Distance Fallen as a Function of Time

The distance fallen by the object as a function of time can be calculated using the equation:

\[
s(t) = s_0 + v_0t + \frac{1}{2}at^2
\]

Given:
- Initial position, \( s_0 = 0 \, \text{m} \) (the object starts from rest)
- Initial speed, \( v_0 = 0 \, \text{m/s} \)
- Acceleration, \( a = g = 9.81 \, \text{m/s}^2 \)

Substituting these values into the equation:

\[
s(t) = 0 + 0 \cdot t + \frac{1}{2}(9.81 \, \text{m/s}^2)t^2 = 4.905t^2 \, \text{m}
\]

The distance graph will be a parabola starting from the origin, opening upwards. At \( t = 5.00 \, \text{s} \), the distance fallen will be:

\[
s(5.00) = 4.905 \times (5.00)^2 = 4.905 \times 25 = 122.625 \, \text{m}
\]

### Summary

1. **Speed graph**: A straight line starting at the origin with a slope of \( 9.81 \, \text{m/s}^2 \).
2. **Distance graph**: A parabola starting at the origin, described by \( s(t) = 4.905t^2 \).

Below are the graphs for the speed and distance as functions of time:

**Graph for Speed vs. Time**: A linear graph starting from (0,0) and increasing with a slope of 9.81. At \( t = 5 \) seconds, speed is approximately 49.05 m/s.

**Graph for Distance vs. Time**: A parabolic graph starting from (0,0) and increasing with the equation \( s(t) = 4.905t^2 \). At \( t = 5 \) seconds, the distance is approximately 122.625 m.

Would you like a detailed sketch of these graphs or have any further questions?

### Follow-Up Questions:

1. How would the graphs change if the object was initially thrown downwards with a certain speed?
2. What if there was air resistance acting on the object? How would the graphs look?
3. How can we calculate the time it takes for the object to reach a certain speed or distance?
4. How would the distance-time graph look for an object thrown upwards and then falling back down?
5. What is the impact of changing gravitational acceleration (e.g., on the Moon) on these graphs?

### Tip:

When dealing with free-fall problems, always start by identifying known quantities (initial speed, time, acceleration) and choose the appropriate kinematic equation to use based on these.

Explore the graphs of speed and distance as functions of time for an object starting from rest under the influence of gravity. Understand the kinematic equations used and the effects of gravity on motion. Suitable for advanced high school students studying physics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation