Answer: 2.7 seconds
Step-by-step explanation:
We only want to answer: How long until the box reaches a height of 13.1 m?
Then we only must integrate the movement equations.
We know that the velocity of the box increases by 3.6 m/s every second, then the acceleration is constant, and can be written as:
a(t) = 3.6m/s^2
Now, for the velocity, we should integrate over time, and because we know that the box starts from rest, the initial velocity (the constant of integration) will be zero.
v(t) = (3.6m/s^2)*t
For the position equation we should integrate again over time, and if we define the position 0 as the ground, we know that the box starts at the ground, then the initial position (the constant of integration) will be zero.
p(t) = (1/2)*(3.6m/s^2)*t^2.
Now we want to find how long will take until the height of the box is equal to 13.1m
Then we must solve:
p(t) = 13.1m = (1/2)*(3.6m/s^2)*t^2
Let's solve this for t.
13.1m = (1/2)*(3.6m/s^2)*t^2
13.1m*2 = (3.6m/s^2)*t^2
26.2m/(3.6m/s^2) = t^2
7.277.... s^2 = t^2
√(7.277.... s^2) = t = 2.7 seconds.
So it will take 2.7 seconds for the box to reach the height of 13.1m