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Sagot :
To solve this problem, we'll use the equations of motion under constant acceleration. The relevant equation here is:
[tex]\[ s = \frac{1}{2} a t^2 \][/tex]
where:
- [tex]\( s \)[/tex] is the distance traveled,
- [tex]\( a \)[/tex] is the acceleration,
- [tex]\( t \)[/tex] is the time.
Given:
- [tex]\( s = 30 \)[/tex] meters,
- [tex]\( a = 2.1 \)[/tex] meters per second squared.
We need to find the time [tex]\( t \)[/tex].
Rearrange the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{2s}{a} \][/tex]
Substitute the given values into the equation:
[tex]\[ t^2 = \frac{2 \times 30 \, \text{m}}{2.1 \, \text{m/s}^2} \][/tex]
[tex]\[ t^2 = \frac{60}{2.1} \][/tex]
[tex]\[ t^2 \approx 28.571 \][/tex]
Now take the square root to solve for [tex]\( t \)[/tex]:
[tex]\[ t \approx \sqrt{28.571} \][/tex]
[tex]\[ t \approx 5.345 \, \text{seconds} \][/tex]
Therefore, the time it takes the hockey player to skate 30 meters is approximately 5.3 seconds.
Thus, the correct answer is:
B. [tex]\( 5.3 \, \text{s} \)[/tex]
[tex]\[ s = \frac{1}{2} a t^2 \][/tex]
where:
- [tex]\( s \)[/tex] is the distance traveled,
- [tex]\( a \)[/tex] is the acceleration,
- [tex]\( t \)[/tex] is the time.
Given:
- [tex]\( s = 30 \)[/tex] meters,
- [tex]\( a = 2.1 \)[/tex] meters per second squared.
We need to find the time [tex]\( t \)[/tex].
Rearrange the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{2s}{a} \][/tex]
Substitute the given values into the equation:
[tex]\[ t^2 = \frac{2 \times 30 \, \text{m}}{2.1 \, \text{m/s}^2} \][/tex]
[tex]\[ t^2 = \frac{60}{2.1} \][/tex]
[tex]\[ t^2 \approx 28.571 \][/tex]
Now take the square root to solve for [tex]\( t \)[/tex]:
[tex]\[ t \approx \sqrt{28.571} \][/tex]
[tex]\[ t \approx 5.345 \, \text{seconds} \][/tex]
Therefore, the time it takes the hockey player to skate 30 meters is approximately 5.3 seconds.
Thus, the correct answer is:
B. [tex]\( 5.3 \, \text{s} \)[/tex]
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