Calculating the distance \( d \) an object travels when initial speed \( v_1 \), acceleration \( a \), and time \( t \) are given is essential in physics and engineering. This article will show you how to determine \( d \) using the distance formula, with detailed steps and examples.
Formula to Find Distance \( d \)
To find the distance \( d \) an object travels under constant acceleration, we use the equation:
\[ d = v_1 \cdot t + \dfrac{1}{2} \cdot a \cdot t^2 \]
Where:
- \( d \) is the distance.
- \( v_1 \) is the initial speed.
- \( t \) is the time interval.
- \( a \) is the linear acceleration.
Step-by-Step Calculation
Let's calculate the distance \( d \) with examples to illustrate how to use this formula.
Example 1: Calculate Distance \( d \)
Given:
- Initial speed \( v_1 = 10 \, \text{m/s} \)
- Acceleration \( a = 2 \, \text{m/s}^2 \)
- Time \( t = 5 \, \text{s} \)
Step-by-Step Calculation:
Step 1: Identify the Given Values
Given:
- Initial speed \( v_1 = 10 \, \text{m/s} \)
- Acceleration \( a = 2 \, \text{m/s}^2 \)
- Time \( t = 5 \, \text{s} \)
Step 2: Substitute the Values into the Distance Formula
Using the formula:
\[ d = v_1 \cdot t + \dfrac{1}{2} \cdot a \cdot t^2 \]
Substitute \( v_1 = 10 \, \text{m/s} \), \( a = 2 \, \text{m/s}^2 \), and \( t = 5 \, \text{s} \):
\[ d = 10 \cdot 5 + \dfrac{1}{2} \cdot 2 \cdot 5^2 \]
Step 3: Calculate the Distance Traveled by Initial Speed
Calculate \( v_1 \cdot t \):
\[ v_1 \cdot t = 10 \cdot 5 = 50 \, \text{m} \]
Step 4: Calculate the Distance Contributed by Acceleration
Calculate \( \dfrac{1}{2} \cdot a \cdot t^2 \):
\[ \dfrac{1}{2} \cdot 2 \cdot 5^2 = 1 \cdot 25 = 25 \, \text{m} \]
Step 5: Add Both Distances
Add \( 50 \, \text{m} \) and \( 25 \, \text{m} \):
\[ d = 50 + 25 = 75 \, \text{m} \]
Final Value
The total distance traveled is \( 75 \, \text{m} \).
Example 2:
Let’s break down another calculation for further clarity.
Given:
- Initial speed \( v_1 = 0 \, \text{m/s} \)
- Acceleration \( a = 4 \, \text{m/s}^2 \)
- Time \( t = 3 \, \text{s} \)
Step-by-Step Calculation:
1. Substitute the Given Values into the Formula:
\[ d = v_1 \cdot t + \dfrac{1}{2} \cdot a \cdot t^2 \]
Given \( v_1 = 0 \, \text{m/s} \), \( a = 4 \, \text{m/s}^2 \), and \( t = 3 \, \text{s} \):
\[ d = 0 \cdot 3 + \dfrac{1}{2} \cdot 4 \cdot 3^2 \]
2. Calculate the Distance Traveled by Initial Speed:
Since \( v_1 = 0 \), \( v_1 \cdot t = 0 \).
3. Calculate the Distance Contributed by Acceleration:
\[ \dfrac{1}{2} \cdot a \cdot t^2 = \dfrac{1}{2} \cdot 4 \cdot 9 = 2 \cdot 9 = 18 \, \text{m} \]
4. Sum Up Both Distances:
\[ d = 0 + 18 = 18 \, \text{m} \]
Thus, the distance traveled is \( 18 \, \text{m} \).
Additional Example
Let’s consider another example to further illustrate:
Example 3: Calculate Distance with Non-Zero Initial Speed
Given:
- Initial speed \( v_1 = 5 \, \text{m/s} \)
- Acceleration \( a = 3 \, \text{m/s}^2 \)
- Time \( t = 4 \, \text{s} \)
Calculation:
1. Substitute into the Formula:
\[ d = v_1 \cdot t + \dfrac{1}{2} \cdot a \cdot t^2 \]
Given \( v_1 = 5 \, \text{m/s} \), \( a = 3 \, \text{m/s}^2 \), and \( t = 4 \, \text{s} \):
\[ d = 5 \cdot 4 + \dfrac{1}{2} \cdot 3 \cdot 4^2 \]
2. Calculate the Distance Traveled by Initial Speed:
\[ v_1 \cdot t = 5 \cdot 4 = 20 \, \text{m} \]
3. Calculate the Distance Contributed by Acceleration:
\[ \dfrac{1}{2} \cdot a \cdot t^2 = \dfrac{1}{2} \cdot 3 \cdot 16 = 1.5 \cdot 16 = 24 \, \text{m} \]
4. Sum Up Both Distances:
\[ d = 20 + 24 = 44 \, \text{m} \]
Thus, the distance traveled is \( 44 \, \text{m} \).
Conclusion
Using the formula \( d = v_1 \cdot t + \dfrac{1}{2} \cdot a \cdot t^2 \), you can calculate the distance an object travels when initial speed, acceleration, and time are known. This method is fundamental for analyzing motion in physics, engineering, and various practical applications.
