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Kinematics Physics Question

  • Thread starter dopey
  • Start date
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1. Homework Statement

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 14 s, then the motor stops. The rocket altitude 18 s after launch is 5000 m. You can ignore any effects of air resistance.
(a) What was the rocket's acceleration during the first 14 _______m/s^2?

(b) What is the rocket's speed as it passes through a cloud 5000 m above the ground?
______m/s


2. Homework Equations
Tried using the kinematic equations


3. The Attempt at a Solution

i dont understand how you can apply the kinematic equations when you dont know the acceleration, it just says it was constant. i know whatever acceleration it is, it is going to be minused from the gravitational acceleration, and you can use the t=14 and t=18 and the x=5000m at t=18. My teacher told me to draw a graph and interpret it, or use the kinematic equations multiple times. Any help would be greatly appreciated.
 

Answers and Replies

Doc Al
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or use the kinematic equations multiple times.
That's what I'd do.

Hint: Call the unknown acceleration "a". Now use the kinematic equations to find expressions for the distance travel during both parts of the motion in terms of "a". (What's the speed of the rocket at the end of the first part of the motion in terms of "a"?) You know that the total distance must add to 5000 m; use that fact to solve for "a".

(Since this is a physics question, not a calculus question, I will move this thread.)
 
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ya sorry, well im in calc based physics and u can apply calculus to this problem by drawing a graph and interpreting the area under the curve (integral) or slope of the line (derivative), so i was seeking some input in that direction.

As for apply the kinematice equation of x=v(initial)t+.5at^2, for the first part, you will have more than one variable. As you said put the speed in terms of "a", are you implying to solve for another equation and substitute, just a little more elaboration would be greatly appreciated.
 
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sorry bout putting it in calculus, i was thinking it was calculus based physics i didnt read underneath it, my apologies
 
Doc Al
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As for apply the kinematice equation of x=v(initial)t+.5at^2, for the first part, you will have more than one variable. As you said put the speed in terms of "a", are you implying to solve for another equation and substitute, just a little more elaboration would be greatly appreciated.
For the first part of the motion, you can assume that the rocket is launched with an initial speed of 0. So the only unknown is the acceleration.

Now try and figure out an expression for the distance traveled during the second part of the motion. Hint: Find the speed at t = 14 s in terms of "a".
 
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even assuming that velocity initial is 0, if u plug it into any equation you still dont know (delta)x or the final velocity so there will be 2 unknowns.

for the second part, trying to solve for the speed in terms of "a" using the final velocity= initial velocity+a(delta)t would be 14a, and do i plug that in the the (delta)x kinematic formula and solve for "a", then subtract gravity, or do u express 5000-(delta)x to give u that height at t=14, ive tried everything ive only got 1 more shot at it.
 
Doc Al
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even assuming that velocity initial is 0, if u plug it into any equation you still dont know (delta)x or the final velocity so there will be 2 unknowns.
So far, for the first part you have:
[tex]x_1 = 1/2 a t^2 = 1/2 a (14)^2[/tex]

You'll be combining x_1 with x_2 (the distance traveled in the second part of the motion), so their sum is a known quantity.

for the second part, trying to solve for the speed in terms of "a" using the final velocity= initial velocity+a(delta)t would be 14a, and do i plug that in the the (delta)x kinematic formula and solve for "a", then subtract gravity, or do u express 5000-(delta)x to give u that height at t=14, ive tried everything ive only got 1 more shot at it.
You found the speed at t = 14s. Good! Hint: Find the speed at t = 18s and use it to find the average speed for part 2. Then you can get an expression for x_2 in terms of "a".

Then use:
[tex]x_1 + x_2 = 5000[/tex]
 

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