Mastering Physics Problem

[Zirka]

Hey. I'm taking first year physics this summer, and one of the components of the course is Mastering Physics (which is loads of fun :yuck:).

I completed the first assignment with a relatively good mark, save for one question which I could not figure out to save my life. There were no hints, so I randomly guessed my four guesses and achieved an amazing 0%. When the answer was displayed, I still couldn't figure out how they managed to get the answer.

This is the question:

Your school science club has devised a special event for homecoming. You've attached a rocket to the rear of a small car that has been decorated in the blue-and-gold school colors. The rocket provides a constant acceleration for 9.0s. As the rocket shuts off, a parachute opens and slows the car at a rate of 5.0 m/s. The car passes the judges' box in the center of the grandstand, 990 m from the starting line, exactly 12 s after you fire the rocket.

What is the car's speed as it passes the judges?

The answer given is 120 m/s, and they provided no explanation.

If someone could explain how to do this problem, I would be very grateful, seeing as the term test is tomorrow. :scared:

[Aronnax]

position = 1/2 a t^2 + (initial velocity) t + (initial position)
velocity = at + (initial velocity)

@ 9s:
position = 81/2 a
velocity = 9 a

from 9-12 seconds acceleration = 0, term falls out
@ 12 seconds:

position = 990m = (position@9s) + (velocity@9s - 5 m/s)t = (81/2)a + 9at - 5t

12s-9s = 3s
t=3

990 = (81/2)a + 27a - 15

a = (1005)/[(81/2)+27] = 14.8...

V@9s = at = (14.8...)(9) = 134 m/s

V@12s = V@9s - 5(t) = 134 - 15 = 119 m/s => rounded => 120 m/s

[zoophilia]

position = 990m = (position@9s) + (velocity@9s - 5 m/s)t = (81/2)a + 9at - 5t

no d = di + vi(t) + 1/2a(t^2)

if you use that you get exactly 15

[Zirka]

Thanks so much Aronnax! Your explanation make perfect sense.

[Aronnax]

no d = di + vi(t) + 1/2a(t^2)

if you use that you get exactly 15

I did, I assumed that acceleration was 0 when I solved for position @9s then put it back in when I solved for velocity.


Correcting my mistake:

position = 1/2 a t^2 + (initial velocity) t + (initial position)
velocity = at + (initial velocity)

@ 9s:
position = 81/2 a
velocity = 9 a

from 9-12 seconds acceleration = -5
@ 12 seconds:

position = 990m = (position@9s) + (velocity@9s)t - (5/2) t^2

12s-9s = 3s
t=3

990 = (81/2)a + 27a - 45/2

a = (1012.5/67.5) = 15

V@9s = at = (15)(9) = 135 m/s

V@9s - 5(3s) = 120 m/s

[zoophilia]

woops, i didn't see your edit

[Zirka]

Thanks zoophilia. Every bit helps, and I'll need all the help I can get, if the test I just wrote was any indication. :scared:

[Aronnax]

woops, i didn't see your edit


When I originally read the problem I took the -5 m/s to be a velocity because of the units and how you can approximate a parachute for short periods of time (to evaluate the acceleration from a parachute it'd be a function of v, not a constant). So at 9s I set:

a = 0
v = (velocity@9s - 5 m/s)

The writers intended for it to be an acceleration and that's what gets you to 120 m/s at 990m. So they wanted:

a= -5
v= velocity@9s

That's what you get for reading a problem too fast:embarassed: