Calculus: Differentiation

In summary: So your answer is correct. In summary, we found the average velocity on the interval [1,2] to be -32 ft/sec, the instantaneous velocity at t=1 and t=2 to be -32 ft/sec and -64 ft/sec respectively, and the time it takes for the object to hit the ground to be approximately 9.2 seconds. We also found the acceleration of the object at any given point in time to be -32 ft/sec^2 and the velocity of the object when it hits the ground to be -294.4 ft/sec.
  • #1
domyy
196
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Homework Statement



The height "s" at time of a silver dollar dropped from a building is given by s(t) = -16t^2 + 1350, where "s" is measured in feet and "t" is measured in seconds
[s'(t) = -32t]

a) Find the average velocity on the interval [1,2].
( I ALREADY SOLVED)
b) Find the instantaneous velocity when t=1 and t=2. (I ALREADY SOLVED)
c) How long will it take to hit the ground?

Not sure about this one. Here's what I did: t=sqrt of 1350/16 ≈ 9.2

d) Find the acceleration of this object at any given point in time.


Not sure about this one. Here's what I did:
v(t) = s'(t) = -32t.

e) What's the velocity of this object when it hits the ground?
(Really have no idea how to find it)

Thanks in advance! I really need to learn how to do this type of problem for my upcoming quiz.
 
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  • #2
domyy said:

Homework Statement



The height "s" at time of a silver dollar dropped from a building is given by s(t) = -16t^2 + 1350, where "s" is measured in feet and "t" is measured in seconds
[s'(t) = -32t]

a) Find the average velocity on the interval [1,2].
( I ALREADY SOLVED)
b) Find the instantaneous velocity when t=1 and t=2. (I ALREADY SOLVED)
c) How long will it take to hit the ground?

Not sure about this one. Here's what I did: t=sqrt of 1350/16 ≈ 9.2
Yes, this is right. Can you explain how you got it?

d) Find the acceleration of this object at any given point in time.


Not sure about this one. Here's what I did:
v(t) = s'(t) = -32t.
OK, that's the velocity as a function of time. How are acceleration and velocity related?

e) What's the velocity of this object when it hits the ground?
(Really have no idea how to find it)
In part (c), you found the time at which the object hits the ground. And you also have the velocity as a function of time. So how can you combine these facts?
 
  • #3
For d) Find the acceleration of this object at any given point in time.
I simply followed what the book states:

"If s is the position function for an object moving along a straight line, then the
acceleration of the object at time t is given by a(t) = v' (t)"

Here's my calculation:
V'(t) = -16(t+Δt)^2 + 1350 - (-16t^2 + 1350)/ Δt

the answer was -32t.

Thank you for assisting me!
 
Last edited:
  • #4
e) what's the velocity of the object when it hits the ground?

"In part (c), you found the time at which the object hits the ground. And you also have the velocity as a function of time. So how can you combine these facts?"

Could it be time x velocity = (9.2)(-32) = 294.4ft/sec ?

Thank you for helping :)
 
  • #5
domyy said:
For d) Find the acceleration of this object at any given point in time.
I simply followed what the book states:

"If s is the position function for an object moving along a straight line, then the
acceleration of the object at time t is given by a(t) = v' (t)"

Here's my calculation:
V'(t) = -16(t+Δt)^2 + 1350 - (-16t^2 + 1350)/ Δt

the answer was -32t.

Thank you for assisting me!
But earlier, you said that v(t) = s'(t) = -32t, which is correct.
Now what is a(t) = v'(t)? You have to differentiate again.
 
  • #6
domyy said:
e) what's the velocity of the object when it hits the ground?

"In part (c), you found the time at which the object hits the ground. And you also have the velocity as a function of time. So how can you combine these facts?"

Could it be time x velocity = (9.2)(-32) = 294.4ft/sec ?

Thank you for helping :)
Well, the answer is right but it's not "time x velocity". It is velocity AT time 9.2, which is v(9.2). And since v(t) = -32t, it follows that v(9.2) = -32*9.2.
 

What is differentiation in calculus?

Differentiation is a mathematical process that involves finding the rate of change of a function at a specific point. It is used to calculate the slope of a curve at a given point, and is an important tool in understanding the behavior of functions.

What is the difference between derivative and differentiation?

The derivative of a function is the result of the differentiation process. In other words, differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of the function at a specific point.

How is the derivative of a function defined?

The derivative of a function is defined as the limit of the ratio of the change in the function's output to the change in its input, as the change in input approaches zero. In other words, it is the slope of the tangent line to the function at a specific point.

What are the basic rules of differentiation?

The basic rules of differentiation include the power rule, product rule, quotient rule, chain rule, and the sum and difference rules. These rules allow us to find the derivatives of more complex functions using the derivatives of simpler functions.

What are some real-world applications of differentiation?

Differentiation has many real-world applications, including in physics, economics, engineering, and other fields. It can be used to calculate rates of change, optimize functions, and model various real-world phenomena such as population growth, velocity, and acceleration.

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