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If a hockey player starts from rest and accelerates at a rate of [tex]2.1 \, m/s^2[/tex], how long does it take him to skate [tex]30 \, m[/tex]?

A. [tex]14.3 \, s[/tex]
B. [tex]5.3 \, s[/tex]
C. [tex]9.8 \, s[/tex]
D. [tex]28.6 \, s[/tex]

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]