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Calculus: Differentiation

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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.
 

Answers and Replies

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jbunniii
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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
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
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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:
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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
jbunniii
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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
jbunniii
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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.
 

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