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A rocket carrying a satellite is accelerating straight up from the Earth's surface. At 1.15 s after
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Jan 13, 2025
A rocket carrying a satellite is accelerating straight up from the Earth's surface. • At 1.15 s after liftoff, the rocket clears the top of its launch platform, which is 63 m above the ground. • After an additional 4.75 s, the rocket is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for: (a) The 4.75-s part of its flight. (b) The first 5.90 s of its flight.
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0:01
a rocket carrying a satellite
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accelerating straight off from the
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Earth's surface at 1.15 seconds after
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lift off the rocket clears the top of it
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is launched platform which is 63 m above
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the ground after an additional 4.75
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seconds the rocket is
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1.0 km above the
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ground so what this question is telling
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us that the rocket is carrying a
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satellite and it is accelerating upward
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from the Ed surface and at 1.15 seconds
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the height is 63 m above the ground
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therefore when after traveling 1.15
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seconds it is the height
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is the height is 63 m above the
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ground
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additionally up
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an additional
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4.75 seconds this 4.75 plus 1.15 seconds
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now calculate the magnitude of the
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average velocity we are to calculate the
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average velocity of the rocket for a
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4.75 seconds part of it is blight and B
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the first 5.90 seconds of it is light
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remember that this 5.90 seconds we got
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it from additional 4. 75 seconds because
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the 4.75 seconds is added to
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1.19 to give
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5.90 so this is why we have the first
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5.90 seconds of it islight now we are to
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extract our data and the time for 4.75
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seconds the time interval is 4.75
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seconds so the initial position that is
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the plus distance so the height covered
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is 63 M and the final which is X1 and
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the final position which is X2 is 1, M
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the reason why we say it's 1,000 m is
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because it is 1,000 K 1 kilom and 1 kilm
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if you convert it into M it is going to
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give you 1,000
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M now the time for 5.9 seconds which is
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the time interval is 5.90 seconds
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therefore now to get the time the
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for the position the initial position is
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0 m because it started from the Flor
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therefore it started at the Flor since
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it started as blow therefore X not is
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equal to 0 m while the final position X
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is 1,000 M which is the same one kilom
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and is converted into M to give us 1,000
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M therefore as you can see X which is
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the final position or the final distance
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or the final height covered is 1,000 M
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while the X not is 0 m which which is at
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the floor therefore now the formula
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we're going to use the average velocity
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formula is B average isal to DX DT that
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is the distance
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cover the change in distance all over
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the change in time where D xals to the
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final distance minus the initial
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distance and DT is the time interval
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therefore now we are going to
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concentrate on
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4.59 sorry
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4.75 seconds before 4 5.19 therefore at
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4.75 seconds we have the xal to final
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minus initial which is X2 - X1 therefore
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we have 1, - 63 and it is going to give
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us 937 that is at 475 seconds therefore
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now to find the aage velocity we say dx/
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d T the X isal to 937 which is the
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height covered and the T isal 4.75
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second which is the time interval
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therefore the final answer for the
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average velocity at 4.75 seconds is
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19726 m/ seconds and this is what we
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have and reason why we had X2 and X1 is
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because X2 is 1 km and it was given us
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1,000 M and X1 is
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63 uh M which was given from the
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question therefore we now divided our
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the total subtracted X1 from
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X2 now for B which is at 5.9 seconds we
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will first find the total height with
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the total distance covered which is the
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final distance minus the initial
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distance the final distance is kilm
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which is 1,000 M and the initial
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distance remember it started from the
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floor from the ground before it covers
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some height therefore at this point when
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it is at the floor or at the ground we
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now
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see it is at 0 m which is 1us 0 and it
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will give us 1,000 M therefore the
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average velocity is to 1,000 the time
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interval which is 5.9 seconds and this
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will give us
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169 49 m/ seconds if you fres your
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calculator remember that the x is
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the uh change in time or sorry change in
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the distance and DT is the time interval
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now we can see that the final answers
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are the average velocity for the 4.75
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seconds is
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19726 m/ seconds and the average
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velocity for the first 5.90 seconds is
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169.99
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m/ seconds and this is our final answer
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