Question
Asked Jul 3, 2019
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If the formula describing the distance s(in feet) an object travels as a function of time(in seconds) is 

s= 100+160t-16t^2

what is the acceleration of the object when t=4?

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Expert Answer

Step 1

The distance s(in feet) an object travels as a function is s = 100 + 160 t -16t2.

It is known that, the velocity of an object is calculated by the derivative of the distance with respect to t.

ds
V=
dt
d
-(100+160t -16/2
dt
=160-32t
Thus, the velocity of an object is v = 160 - 32t
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Image Transcriptionclose

ds V= dt d -(100+160t -16/2 dt =160-32t Thus, the velocity of an object is v = 160 - 32t

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Step 2

Note that, the acceleration of an object is calculated by the ...

dv
a
dt
-(160-32t)
dt
=-32
Here, the acceleration is constant
Thus, the acceleration when t
4 is -32feet/s2.
help_outline

Image Transcriptionclose

dv a dt -(160-32t) dt =-32 Here, the acceleration is constant Thus, the acceleration when t 4 is -32feet/s2.

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Tagged in

Math

Calculus

Derivative

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