Integration Problem

[dlever]

I was hoping someone could help with this calculus problem.
In a manufacturing process, bags of material on an upper conveyor belt moving at 24.3 cm/s drop onto another conveyor belt 76.2 cm below.
a. If the process requires that the bag drop off the upper belt just as the preceding bag hits the lower belt, how far apart should the bags be placed on the upper belt?
b. If the bags are to be 22.9 cm apart on the lower belt, what is the required speed of the belt?
any help would be greatly appreciated

 

[Deleted member 4993]

I was hoping someone could help with this calculus problem.
In a manufacturing process, bags of material on an upper conveyor belt moving at 24.3 cm/s drop onto another conveyor belt 76.2 cm below.
a. If the process requires that the bag drop off the upper belt just as the preceding bag hits the lower belt, how far apart should the bags be placed on the upper belt?
b. If the bags are to be 22.9 cm apart on the lower belt, what is the required speed of the belt?
any help would be greatly appreciated

Start with defining variables:

Let the spacing on upper conveyor belt = d

How long does it take for the bag to fall from the upper to lower conveyor belt?

Please share your work with us, indicating excatly where you are stuck - so that we may know where to begin to help you.

 

[dlever]

This is what I started with:

76.2 = Distance
76.2 = 1/2gt^2
(2)(76.2)/980 = t^2
0.156 = t

I'm not sure if this is correct since I really haven't created an integral.

 

[Deleted member 4993]

This is what I started with:

76.2 = Distance
76.2 = 1/2gt^2
(2)(76.2)/980 = t^2
0.156 = t

I'm not sure if this is correct since I really haven't created an integral.

You are good.

This is a simple problem - you don't need integral.

Now what .....

 

[dlever]

Subhotosh Khan,
Thanks for the help, but since the assignment revolves around Integration I would need to create an Integral, that's why I am stuck.
To answer the rest of the problem, I get:
a. separation between bags should be (24.3 cm/s)(0.156s) = 3.79 cm
b. lower belt needs to be moving at 22.9cm/0.156 s = 146.8 cm/s

Could you possibly point me in the right direction for an Integral ?

 

[Deleted member 4993]

This is what I started with:

76.2 = Distance
76.2 = 1/2gt^2
(2)(76.2)/980 = t^2 0.156 = t

I'm not sure if this is correct since I really haven't created an integral.

That equation can be written in integral form:

\(\displaystyle s \ = \ \int_{t_1}^{t_2}v(t)\cdot dt\)