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A rock is dropped from a height of 64ft. Its height above ground at time t seconds later is given by s(t)=−16t2+64 for 0≤t≤2. Find its instantaneous velocity 1 second after it is dropped using the limit definition of a derivative. Do not include units in your answer.

Question

A rock is dropped from a height of 64ft. Its height above ground at time t seconds later is given by s(t)=−16t2+64 for 0≤t≤2. Find its instantaneous velocity 1 second after it is dropped using the limit definition of a derivative. Do not include units in your answer.

check_circleAnswer
Step 1

Instantaneous velocity at t=1 is ds/dt at t=1 , that is s'(1).

s(t)16+64
S
s'() lim )(1)
t-1
help_outline

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s(t)16+64 S s'() lim )(1) t-1

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Step 2

We substitute s(t) and s(1) in the lim...

(-16 +64)(-16(1) 64)
s'(1) lim-(lim-
t1
t 1
t-1
-16 (-1)
-16t216
- lim
-16t264 16 64
- lim
= lim
t-1
t-1
t-1
-16(t-1)(t+1) lim[-16(t+1)=-
lim
32
t 1
help_outline

Image Transcriptionclose

(-16 +64)(-16(1) 64) s'(1) lim-(lim- t1 t 1 t-1 -16 (-1) -16t216 - lim -16t264 16 64 - lim = lim t-1 t-1 t-1 -16(t-1)(t+1) lim[-16(t+1)=- lim 32 t 1

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Tagged in

Math

Calculus

Derivative

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