Question

help please. Rocket problem; impulse, change in velocity

Answer 1

at t=0, a 2250kg in outer space fires an engine that exerts an increasing force on it in the +x-direction. This force obeys the equation \[F_{x}=At^2\] where 1.25s is time, and has a magnitude of 781.25N when t=1.25s

1) Find the SI constant A.

2) What impulse does the engine exert on the rocket during the 1.50s interval starting 2.00s after the engine was fired?

3) By how much does the rocket's velocity change during this interval?

Answer 2

i started by just plugging in numbers into the equation. like 781.25 for F(x) and 1.25 for t and figured out the first part of the question. i got 500 N/s^2

Answer 3

I'm a bit lost for finding the impulse.

Answer 4

what is impulse? mv1 - mv2 right?

Answer 5

yeah.

Answer 6

how do we determine the velocity at a given moment in time?

Answer 7

take the integral of an equation?

Answer 8

or was it derivative. i think it was derivative

Answer 9

derivative of postition, or integral of acceleration :)

Answer 10

so right now it's in the form of ma since i have newtons as the result for number 1 right?

Answer 11

i meant N/s^2

Answer 12

if F is a force, then the force exerted at any time parameter is ma yes

ma = At^2

a = At^2/m seems reasonable to me

Answer 13

hmmm ok. let me see what happens.

Answer 14

this is just an idea, ill know if its good if we have an answer key to compare with

Answer 15

ok so for a i get 0.3472 m/s^2

Answer 16

oh i just found in my notes that \[\int\limits_{a}^{b}f(t)dt= \Delta p\]

Answer 17

so i guess i should just take the integral and see what happens.

Answer 18

oh wait. the question is asking for the time interval of 1.50s-2.00s. i guess i started using the wrong time since i used 1.25s right?

Answer 19

ok so for at a(t=1.50)=0.5 m/s^2 and F=1125 and at t=2, f=2000 and a(t=2)=0.89m/s^2

Answer 20

was indisposed for a time

Answer 21

so i guess to find impuls of this interval do i just do this \[\int\limits_{1.50}^{2.00}500t^2dt\]

Answer 22

so then it's just \[\frac{ 500 }{ 3 }(1.50)^3-\frac{ 500 }{ 3 }(2.00)^3\]

Answer 23

and i'm getting -770.83 for this. does this look correct?

Answer 24

oh cripes. it says the interval 1.50s starting 2.00s AFTER the engine was fired.

Answer 25

a = At^2/m

v = At^3/(3m)

mv = At^3/3

Answer 26

so yes

A/3 (t1^3 - t0^3) for the time interval asked for

Answer 27

ok so the time interval is starting at 2 and will end at 2.5 since it said it was an interval of 1.5s and starting at 2 right?

Answer 28

ugh. starting at 2 and ending at 3.5. bad math here....

Answer 29

ok so the answer i'm getting is 5812.5

Answer 30

hey i just entered it into mastering physics and got it! yay. lol

Answer 31

thank you thank you. now i guess change in velocity.

Answer 32

well since impluse = m(v1-v0) then divide the impluse by the mass to get change in velocity :)

Answer 33

haha! so simple! thank you!

Answer 34

and change in velocity is 2.58

Answer 35

i can handle the rest. thank you

Answer 36

youre welcome :)