A rocket travelling at +88m/s is accelerated to +132m/s over a 15s interval. What is its displacement during this time?

Answer 1

The displacement is #=1650m#

Apply the equation of motion

#v=u+at#

The acceleration is

#a=(v-u)/t=(132-88)/15=2.93ms^-2#

The displacement is calculated with the equation

#v^2=u^2+2as#

The displacement is

#s=(v^2-u^2)/(2a)=(132^2-88^2)/(2*2.93)=1650m#
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Answer 2

The displacement of the rocket during this time is 2100 meters.

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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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