Answer :
To solve this problem, we can use one of the equations of motion for uniformly accelerated motion. The equation we will use is:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Where:
- [tex]\( s \)[/tex] is the displacement
- [tex]\( u \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration
- [tex]\( t \)[/tex] is the time
Given:
- Acceleration, [tex]\( a = 10 \, \text{m/s}^2 \)[/tex]
- Initial velocity, [tex]\( u = 0 \, \text{m/s} \)[/tex] (since it starts from rest)
- Time, [tex]\( t = 5 \, \text{seconds} \)[/tex]
Let's plug these values into the equation:
1. First, calculate [tex]\( ut \)[/tex]:
[tex]\[ ut = 0 \times 5 = 0 \][/tex]
2. Next, calculate [tex]\( \frac{1}{2}at^2 \)[/tex]:
[tex]\[ \frac{1}{2} \times 10 \times (5)^2 = \frac{1}{2} \times 10 \times 25 = 5 \times 25 = 125 \, \text{m} \][/tex]
Now, add these together to find the total displacement:
[tex]\[ s = 0 + 125 = 125 \, \text{meters} \][/tex]
Therefore, the displacement of the object after 5 seconds is [tex]\( 125 \)[/tex] meters.
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Where:
- [tex]\( s \)[/tex] is the displacement
- [tex]\( u \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration
- [tex]\( t \)[/tex] is the time
Given:
- Acceleration, [tex]\( a = 10 \, \text{m/s}^2 \)[/tex]
- Initial velocity, [tex]\( u = 0 \, \text{m/s} \)[/tex] (since it starts from rest)
- Time, [tex]\( t = 5 \, \text{seconds} \)[/tex]
Let's plug these values into the equation:
1. First, calculate [tex]\( ut \)[/tex]:
[tex]\[ ut = 0 \times 5 = 0 \][/tex]
2. Next, calculate [tex]\( \frac{1}{2}at^2 \)[/tex]:
[tex]\[ \frac{1}{2} \times 10 \times (5)^2 = \frac{1}{2} \times 10 \times 25 = 5 \times 25 = 125 \, \text{m} \][/tex]
Now, add these together to find the total displacement:
[tex]\[ s = 0 + 125 = 125 \, \text{meters} \][/tex]
Therefore, the displacement of the object after 5 seconds is [tex]\( 125 \)[/tex] meters.
