When calculating the time it takes to cover a distance, we can use the formula:
\[ d = v_1 \cdot t + \dfrac{1}{2} \cdot a \cdot t^2 \]
Where:
- \( d \) is the distance
- \( v_1 \) is the initial velocity
- \( a \) is the acceleration
- \( t \) is the time
In this article, we'll explore three scenarios:
1. The initial speed is zero.
2. The acceleration is zero.
3. The acceleration is negative (e.g., a car braking).
Finding Time When Initial Speed is Zero
Formula Simplification
When the initial speed \( v_1 \) is zero, the formula simplifies to:
\[ d = \dfrac{1}{2} \cdot a \cdot t^2 \]
Solving for \( t \):
\[ t = \sqrt{\dfrac{2d}{a}} \]
Example 1: A Car Accelerating from Rest
Scenario: A car accelerates from rest at 2 m/s² and covers a distance of 100 meters.
- Distance (\( d \)) = 100 meters
- Acceleration (\( a \)) = 2 m/s²
Calculation:
\[ t = \sqrt{\dfrac{2 \cdot 100}{2}} = \sqrt{100} = 10 \text{ seconds} \]
Finding Time When Acceleration is Zero
When the acceleration \( a \) is zero, the formula simplifies to:
\[ d = v_1 \cdot t \]
Solving for \( t \):
\[ t = \dfrac{d}{v_1} \]
Example 2: A Cyclist Traveling at Constant Speed
Scenario: A cyclist travels at a constant speed of 5 m/s and covers a distance of 50 meters.
- Distance (\( d \)) = 50 meters
- Initial speed (\( v_1 \)) = 5 m/s
Calculation:
\[ t = \dfrac{50}{5} = 10 \text{ seconds} \]
Finding Time When Acceleration is Negative (Braking)
When a car is braking, the acceleration is negative. Let's solve for \( t \) in this situation:
\[ d = v_1 \cdot t + \dfrac{1}{2} \cdot (-a) \cdot t^2 \]
This can be rearranged to:
\[ 0 = \dfrac{1}{2} \cdot a \cdot t^2 - v_1 \cdot t + d \]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where:
- \( a = \dfrac{1}{2} \cdot a \)
- \( b = -v_1 \)
- \( c = d \)
Solving this quadratic equation for \( t \):
\[ t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Example 3: A Car Braking to a Stop
Scenario: A car traveling at 20 m/s begins to brake with an acceleration of -4 m/s² and needs to cover 50 meters before stopping.
- Distance (\( d \)) = 50 meters
- Initial speed (\( v_1 \)) = 20 m/s
- Acceleration (\( a \)) = -4 m/s²
Calculation:
Using the quadratic formula:
\[ 0 = \dfrac{1}{2} \cdot (-4) \cdot t^2 - 20 \cdot t + 50 \]
Simplifying:
\[ 0 = -2t^2 - 20t + 50 \]
Solving for \( t \):
\[ t = \dfrac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot (-2) \cdot 50}}{2 \cdot (-2)} \]
\[ t = \dfrac{20 \pm \sqrt{400 + 400}}{-4} \]
\[ t = \dfrac{20 \pm \sqrt{800}}{-4} \]
\[ t = \dfrac{20 \pm 20\sqrt{2}}{-4} \]
Taking the positive solution:
\[ t = \dfrac{20(1 + \sqrt{2})}{-4} = 5(1 + \sqrt{2}) \approx 12.07 \text{ seconds} \]
These examples illustrate how to find the time it takes to cover a distance in different scenarios using the provided formula. Whether you're accelerating from rest, traveling at a constant speed, or braking, this approach helps you determine the time required to cover a specified distance.
