How do you find the acceleration for the function #s(t)=-16t^2+100t +20# ?

Answer 1

# -32" unit."#

The velocity #v# of the particle travelling distance #s# in time #t# is given
by, #v=(ds)/dt#.
The acceleration #a# then is #a=(dv)/dt=(d^2s)/dt^2#.
#:. v=(ds)/dt=d/dt{s(t)}#
#=d/dt{-16t^2+100t+20}#
#=-16*2t+100#.
# rArr v=-32t+100#.
Then, #a=(dv)/dt=d/dt{-32t+100}=-32" unit."#
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Answer 2

To find the acceleration function, you need to take the second derivative of the position function ( s(t) ) with respect to time (( t )). The position function given is ( s(t) = -16t^2 + 100t + 20 ). Taking the second derivative of ( s(t) ) with respect to ( t ) yields the acceleration function, denoted as ( a(t) ).

[ s(t) = -16t^2 + 100t + 20 ] [ v(t) = \frac{ds}{dt} = \frac{d}{dt}(-16t^2 + 100t + 20) ] [ a(t) = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{ds}{dt}\right) = \frac{d^2s}{dt^2} ]

Now, differentiate ( s(t) ) twice with respect to ( t ) to find ( a(t) ).

[ a(t) = \frac{d}{dt}(-16t^2 + 100t + 20) ] [ a(t) = -32t + 100 ]

So, the acceleration function for the given position function is ( a(t) = -32t + 100 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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