To determine the distance traveled by the object in the 3rd second, we need to consider the motion of the object and the time it takes to reach that point.
When an object is thrown vertically upward, its initial velocity is positive (upward) and gravity acts in the opposite direction, causing it to decelerate. At the highest point, the object momentarily comes to rest and then starts to fall back down.
To find the distance traveled in the 3rd second, we first calculate the time it takes for the object to reach its highest point. Since the object is thrown upward, its velocity decreases at a constant rate of approximately 9.8 m/s² due to gravity.
Using the equation for velocity in free fall:
v = u - gt,
where: v = final velocity (0 m/s at the highest point), u = initial velocity (25 m/s), g = acceleration due to gravity (-9.8 m/s²), t = time taken.
Rearranging the equation, we have:
t = (v - u) / g.
Substituting the given values:
t = (0 - 25) / (-9.8) = 2.55 seconds (rounded to two decimal places).
So, it takes approximately 2.55 seconds for the object to reach its highest point. The object will then take the same amount of time to fall back down, resulting in a total time of approximately 2.55 seconds for the entire upward and downward journey.
Now, we can find the distance traveled in the 3rd second. Since the first second corresponds to t = 0 to t = 1 second, and the second second corresponds to t = 1 to t = 2 seconds, the third second corresponds to t = 2 to t = 3 seconds.
During this time interval, the object is moving downward due to gravity. The velocity increases at a rate of approximately 9.8 m/s². Thus, we can use the formula for distance traveled under constant acceleration:
s = ut + (1/2)at²,
where: s = distance traveled, u = initial velocity (0 m/s at the highest point), t = time taken (3 seconds), a = acceleration (-9.8 m/s²).
Substituting the values:
s = 0(3) + (1/2)(-9.8)(3)² = -44.1 meters.
Since distance cannot be negative in this context, we take the magnitude of the result: |s| = 44.1 meters.
Therefore, the object travels a distance of approximately 44.1 meters during the 3rd second.