If a #2 kg# object is constantly accelerated from #0m/s# to #15 m/s# over # 12s#, how much power musy be applied at #t= 3 #?

Answer 1

9.4W

Constant acceleration here denotes a constant force #F.#
To find out the power #P# required, we need to find out #(dW)/(dt)# , the rate of work done with respect to time at #t=3s#. Alternatively, the equation #P=Fv# can be used.
#P=(dW)/(dt)=(d(Fx))/(dt)=F((dx)/(dt))=Fv=mav#
To find #a# , we can use the kinematics equation #a=(v-u)/t#.
#a=(15-0)/12=1.25m/s^2#
To find #v# at #t=3s# , we can use #v=u+at#.
#v=0+(1.25)(3)=3.75m/s#

Therefore, the power required is

#P =mav =(2)(1.25)(3.75) =9.375 =9.4W#

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Answer 2

The power that must be applied at t = 3s is 10 Watts.

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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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