Answer:
Explanation:
The equation of equlibrium for the box is:
The formula for the acceleration, given in 
, is:
Velocity can be derived from the following definition of acceleration:
![v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B2%5Ccdot%5B%282.278%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%20%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D-%280.034%5C%2C%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%5E%7B2%7D%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D%5D%20%20%7D)
The speed after the box has travelled 17 meters is:
By definition, acceleration is the change in velocity per change of time. As time passes by, the time increases in value. So, when the acceleration is decreasing while the time is increasing, then that means that the change of velocity is also decreasing with time. So, optimally, the initial velocity and the velocity at any time are very relatively close to each other,
I'm not sure if this is correct but it's what I'll do
This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.
Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )
Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12
Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2
4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t
Time for Stone B is 4s
Time for Stone A is 6s
