Acceleration problem

  • Thread starter ryanbersa
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  • #1
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Homework Statement



a rocket is launched with constant acceleration for 16 secs and after 20 secs its reaches 5100m

what is acceleration during first 16 secs
what is velocity at 5100m

Homework Equations



v=v0+at

The Attempt at a Solution



I calculated the avg acceleration as 25.5 m/s^2
I calculated the avg velocity as 255 m/s^2

I know after 16 seconds the acceleration is -9.8 m/s^2

i have tabulated various quants to try make sense of it and even did a position vs time graph but i can get further

also did motion diagrams to show velocity vector and acceleration vectors
 

Answers and Replies

  • #2
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Suppose you call the acceleration during the first 16 seconds "a". What is the velocity of the rocket at time zero (as it is being launched)? In terms of "a", what is its altitude after the first 16 seconds, and what is its velocity after the first 16 seconds?

Chet
 
  • #3
Doc Al
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How could you express (mathematically) the distance traveled during the first 16 seconds in terms of the unknown acceleration? (Call the acceleration "a".)

How could you express (mathematically) the distance traveled during the next 4 seconds?

What must they add up to?

Edit: Chestermiller beat me to it!
 
  • #4
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Velocity at start is zero and I don't have any values for at time 16s
 
  • #5
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Ok thanks guys I have an idea now so will give it another stab in the morning

Your help is much appreciated :) I managed the first 50 questions of this chapters exercise set but this one got me
 
  • #6
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Hi guys

If I use y1 = y0 + v0(t) + 1/2(a)(t)^2 then I get Y1=128a and Y2=8a

so if i understand Doc Al correct then this should add up to 5100 meters so

128a_1+8a_2=5100m

a_1=(5100-8a_2)/128
 
  • #7
Nathanael
Homework Helper
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If I use y1 = y0 + v0(t) + 1/2(a)(t)^2 then I get Y1=128a and Y2=8a
Why does Y2=8a?

What's the velocity at the beginning of Y2? What's the acceleration?


Edit:
and even did a position vs time graph
Try using a velocity vs time graph. That's actually an excellent way to solve the problem.
(You'll need to know two basic "calculus" ideas: 1.) the slope of the velocity graph is the acceleration, and 2.) the area of the velocity graph is the distance. Since acceleration is constant, the slope is constant, meaning the graph is made of only straight lines.)

If you don't immediately see the solution, then imagining the velocity vs time graph is certainly a great way to jump straight to the final equation.
 
Last edited:
  • #8
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y1 = 128a
v= 16a

then

y2=y1+v1(t)-1/2g(t)^2
5100=128a+16a(4)-1/2(9.8)(4)^2
5178=192a
27m/s=a ?
 
  • #9
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and

Vf=Vi+a(t)
Vf=16a+9.8(4)
Vf=16(27)-9.8(4)
Vf=392.8
 
  • #10
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By trial and error on excel :
Plug in :
initial acceleration for 15 seconds @ 27.85 (m/s)/s
Then decelerate under gravity for 5 seconds

This should satisfy the results
 
  • #11
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dont have excel in the exam :)
 
  • #12
Doc Al
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y1 = 128a
v= 16a

then

y2=y1+v1(t)-1/2g(t)^2
5100=128a+16a(4)-1/2(9.8)(4)^2
5178=192a
27m/s=a ?
Good! (Except the units should be m/s^2.)

and

Vf=Vi+a(t)
Vf=16a+9.8(4)
Vf=16(27)-9.8(4)
Vf=392.8
Good!
 
  • #13
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Phase 1 : accelerating under power for 15 seconds.
Phase 2 : decelerating unpowered for 5 seconds under gravity.

If you get the answer by mathematical means, you will know if your procedure is correct (because you have the answer) saving you further work.
You could post this enquiry on the mathematics section, you may get an improved response.
By the way the velocity at 5100 m is 368.8 m/s exactly, and an improved answer for the acceleration is
27.8533 (m/s)/s with the 3's recurring.

Ive based these calcs on inputting an arbitrary value for (phase 1) acceleration, then calculating the distance and velocity after 15 seconds (when the motor shuts down), this velocity is then used for the slowing by gravity stage (lasting 5 seconds) as the initial velocity.
Newtons equations of motion are used for both stages.
I altered the value for initial acceleration value until the distance was correct ( (trial and error)
The maths algorithm is beyond me, but it must exist or you wouldnt have been asked.
Hope this helps
Dean
 
  • #14
Doc Al
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Phase 1 : accelerating under power for 15 seconds.
Phase 2 : decelerating unpowered for 5 seconds under gravity.
Those phases should be 16 and 4 seconds.

There's no need for trial and error. The math is straightforward, as ryanbersa has demonstrated.
 
  • #15
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oops, not reading the question properly, might as well use the algorithm again :
revised solutions (16 and 4 seconds)
acceleration = 26.9708333 (3's recurring)
final velocity (g at - 9.8 (m/s)/s) = 392.333 (3's recurring)
 

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