Sam is pulling a box up to the second story of his apartment via a string. The box weighs 92.6 kg and starts from rest on the ground. Sam is pulling so that the velocity of the box is increasing by 3.6 m/s every second. How long until the box reaches a height of 13.1 m?

2.7 seconds

Explanation:

We only want to 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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14