Answer:
The correct option 50.0 m (the units of the answers are wrong, they should be in meters and not in millimeters).
Explanation:
From the first second we can find the acceleration of the object:
[tex] X_{f} = X_{0} + V_{0}t + \frac{1}{2}at^{2} [/tex]
Where:
[tex]X_{f}[/tex]: is the final position = 2.00 m
[tex]X_{0}[/tex]: is the initial position = 0
[tex]V_{0}[/tex]: is the initial speed = 0 (starts from rest)
t: is the time = 1 s
The acceleration is:
[tex] a = \frac{2X_{f}}{t^{2}} = \frac{4.00 m}{(1 s)^{2}} = 4.00 m/s^{2} [/tex]
Now, we can find the distance after 5 seconds:
[tex]X_{f} = X_{0} + V_{0}t + \frac{1}{2}at^{2} = \frac{1}{2}*4.00 m/s^{2}*(5 s)^{2} = 50.0 m[/tex]
Therefore, the correct answer is 50.0 m. All the options have the units in millimeters and not in meters hence, the units of the options are wrong.
I hope it helps you!
The final position of the object after 5 seconds is 50 m.
Option D is the correct answer.
Given that the object starts from rest and accelerates uniformly. It moves 2.00m during the first second.
Let us consider that the initial velocity is v and the initial position is x and the final position is x' for time interval t.
Hence for the given condition, t = 1 s, x' = 2 m, v = 0 m/s and x = 0 m. The acceleration can be calculated as given below.
[tex]a = \dfrac {2x'}{t^2}[/tex]
[tex]a = \dfrac {2\times 2}{1}[/tex]
[tex]a = 4 \;\rm m/s^2[/tex]
The final position of the object after 5 seconds is given below.
[tex]x (\rm final) = x + vt + \dfrac {1}{2}at^2[/tex]
Substituting the values in the above equation.
[tex]x(\rm final) = 0 + 0 + \dfrac {1}{2}\times 4 \times 5^2[/tex]
[tex]x (\rm final) = 50 \;\rm m[/tex]
Hence we can conclude that the final position of the object after 5 seconds is 50 m. Option D is the correct answer.
To know more about acceleration and velocity, follow the link given below.
https://brainly.com/question/2239252.
