Answer :
Sure! Let's solve for the displacement \( \Omega \) using the given formula:
[tex]\[ \Omega = u t + \frac{1}{2} a t^2 \][/tex]
Given values:
- Initial velocity, \( u = 10 \) meters per second (m/s)
- Time, \( t = 5 \) seconds (s)
- Acceleration, \( a = 2 \) meters per second squared (m/s²)
Now, let's substitute these values into the formula step-by-step.
1. Calculate the first term \( u t \):
[tex]\[ u t = 10 \, \text{m/s} \times 5 \, \text{s} = 50 \, \text{m} \][/tex]
2. Calculate the second term \( \frac{1}{2} a t^2 \):
[tex]\[ \frac{1}{2} a t^2 = \frac{1}{2} \times 2 \, \text{m/s}^2 \times (5 \, \text{s})^2 \][/tex]
Simplify inside the parentheses first:
[tex]\[ \frac{1}{2} \times 2 \, \text{m/s}^2 \times 25 \, \text{s}^2 = 1 \times 25 \, \text{m} = 25 \, \text{m} \][/tex]
3. Add the two terms together:
[tex]\[ \Omega = 50 \, \text{m} + 25 \, \text{m} = 75 \, \text{m} \][/tex]
Therefore, the displacement [tex]\( \Omega \)[/tex] is [tex]\( 75 \)[/tex] meters.
[tex]\[ \Omega = u t + \frac{1}{2} a t^2 \][/tex]
Given values:
- Initial velocity, \( u = 10 \) meters per second (m/s)
- Time, \( t = 5 \) seconds (s)
- Acceleration, \( a = 2 \) meters per second squared (m/s²)
Now, let's substitute these values into the formula step-by-step.
1. Calculate the first term \( u t \):
[tex]\[ u t = 10 \, \text{m/s} \times 5 \, \text{s} = 50 \, \text{m} \][/tex]
2. Calculate the second term \( \frac{1}{2} a t^2 \):
[tex]\[ \frac{1}{2} a t^2 = \frac{1}{2} \times 2 \, \text{m/s}^2 \times (5 \, \text{s})^2 \][/tex]
Simplify inside the parentheses first:
[tex]\[ \frac{1}{2} \times 2 \, \text{m/s}^2 \times 25 \, \text{s}^2 = 1 \times 25 \, \text{m} = 25 \, \text{m} \][/tex]
3. Add the two terms together:
[tex]\[ \Omega = 50 \, \text{m} + 25 \, \text{m} = 75 \, \text{m} \][/tex]
Therefore, the displacement [tex]\( \Omega \)[/tex] is [tex]\( 75 \)[/tex] meters.
