(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known dataa 1000kg weather rocket is launched straight up, the rocket motor provides a constant accerelation for 16s, then the motor stops. the rocket altitude 20s after launch is 5100m you can ignore any effects of air resistance.

what is the rocket's acceleration during the first 16s

2. Relevant equations

d=[Vf^2-Vi^2]/2a

delta x= xi+viT+1/2aT^2

vf-vi=aT

3. The attempt at a solution

just checking my answer since there isn't one in the back of the book.. so this is what i did

1. i first found the distance of the 4s after the motor stop

t=4s a=-9.8m/s^2 Vf=0

so i found Vi

vf-vi=aT

0-Vi=-9.8(4)

Vi=39.2m/s

2. then i use thosenumbers and plugged it into the 1st equation

d=[vf^2-vi^2]/2a

d=0-39.2/[2*-9.8]

d=2

therefore the total distance travelled in the first 16s is 5100-2m = 5098m

3. then i took the 2nd equation and plugged in that to find the accleration

5098=0+0+1/2a(16^2)

a=39.83m/s^2

Did i get the right answer?

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# Homework Help: A rocket is launched kinematics

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