Am i wrong or the book?

  • Thread starter elpermic
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  • #1
elpermic
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Homework Statement


A rocket carrying a satellite is accelerating straight up from the Earth's surface. At 1.35 s after liftoff, the rocket clears the top of its launch platform. After an additional 4.45s it is 1.00km above the ground. Calculate the magnitude of the average velocity of the rocket for a) the 4.45-s part of its flight; b) the first 5.80s of its flight.


Homework Equations





The Attempt at a Solution



A)I did a and kept on getting 225. (1063-63)/(5.80-1.35)=225
B) I don't get B
 

Answers and Replies

  • #2
willem2
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The problem isn't very clear

do you have to assume constant accelaration for some part or all of the flight?

the altitude of the launch platfrom is 63m?

the rocket is at 1 km above the ground after 5.8s. part B seems simple.
 
  • #3
Delphi51
Homework Helper
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Where did the 63 come from? Is this the height of the launch platform and was it given?
You use v = d/t, or d = vt, which is not correct for accelerated motion.
Rather d = Vi*t + .5*a*t^2 should be used (assuming acceleration is constant).

My method was to use the letter d for the vertical distance traveled in the first 1.35s.
Using that v = formula for accelerated motion on the first 1.35 s of flight, I got a relationship between d and the acceleration.
Using the formula again for the next 4.45 s, I got another relationship between d and a.
Combining the two, I was able to find both d and a. I did not get your 63 m for d.
Having these values made it easy to find the velocities at times 0, 1.35 and 5.8, so simply adding two of them and dividing by 2 gave the average velocity in any interval. I got the 225 m/s value for the 4.45 s interval.
 
  • #4
elpermic
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63 is the top of the launch platform
 
  • #5
Delphi51
Homework Helper
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Knowing the 63 m at time 1.35 makes a big difference in the question!
My assumption of constant acceleration is wrong - you can easily calculate the acceleration over the two time intervals and you get different answers.
Indeed, rockets increase their acceleration as they burn off fuel and decrease their mass.

The question only asks for average velocities and the simple formula for average velocity is just V = (change in distance)/(change in time).
That is what you did for the 4.45 s: (1063-63)/(5.80-1.35)=225
but shouldn't it be (1000-63)/4.45 = 211 ?

Do the (b) part the same way using the 63 meters in time 1.35 s.
 
  • #6
D H
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A)I did a and kept on getting 225. (1063-63)/(5.80-1.35)=225
Whether you gave us the full question is irrelevant. I can tell upon inspection that your answer is incorrect. Why? You answer is not an acceleration. Get into the habit of always being explicit about units. The cost in doing so is minimal: you have to write a few extra characters. The benefit is immense. You will know whether you answer makes sense from a units perspective. If you don't have the units right you don't have the right answer.

In this case,

[tex](1063\,\text{m}\,-63\,\text{m})/(5.80\,\text{s}\,-1.35\,\text{s}) = 225\,\text{m}/\text{s}[/tex]

You have just computed a speed, not an acceleration.


BTW, that 1063 meters is also incorrect. The vehicle is 1 km above the ground after 5.80 seconds, not 1 km above the top of the launch platform.
 
  • #7
willem2
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You have just computed a speed, not an acceleration.

the question was to compute a speed
 
  • #8
D H
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You're right. I misread the question.
 

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