If a baseball has a zero velocity at some instant, is the acceleration of the baseball necessarily zero at that time? Explain -
1 answer:
Answer:
No, not necessarily
Explanation:
If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.
An example of an object that has zero velocity but non-zero acceleration:
If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8 rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8
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Answer:
Explanation:
For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).
With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:
γhxdy
we know that γ=62.5 lb/
from the drawing, we can determine the height (or depth under the water) of each differential area is given by:
h=8-y
x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:
when substituting the x and y-values given on the graph, we get that the slope is:
once we got this slope, we can substitute it in the point-slope form of the equation:
which yields:
which simplifies to:
we can now solve this equation for x, so we get that:
with this last equation, we can substitute everything into our integral, so it will now look like this:
Now that it's all written in terms of y we can now simplify it, so we get:
we can now proceed and evaluate it.
When using the power rule on each of the terms, we get the integral to be:
By using the fundamental theorem of calculus we get:
When solving we get:
