How Far Does a Rocket Sled Travel with Variable Acceleration?

  • Thread starter Thread starter eluu
  • Start date Start date
  • Tags Tags
    Dynamics Motion
Click For Summary
The discussion revolves around calculating the distance traveled by a rocket sled with variable acceleration. Initially, the sled accelerates at a rate of a = 30 + 2t until it reaches a velocity of 400 m/s, with the first calculation yielding a displacement of 1833.333 meters after correcting the time to 10 seconds. Following this, the sled experiences deceleration due to a water brake, where the acceleration is defined as a = -0.003v^2, leading to further calculations. The total distance traveled, including both acceleration and deceleration phases, was initially miscalculated, but corrections were made to arrive at a total distance of approximately 2163.750 meters, though the expected answer is 2300 meters. The discussion highlights the importance of using calculus for variable acceleration scenarios.
eluu
Messages
9
Reaction score
0
[SOLVED] Motion Question (Dynamics)

EDIT: I just realized this question was better suited in this forum than the engineering one, sorry :(



Homework Statement



The rocket sled starts from rest and accelerates at a = 30 + 2t until its velocity is 400m/s. It then hits a water brake and its acceleration is a = -0.003v^2 m/s^2 until its velocity decreases to 100 m/s. What total distance does it travel?

Homework Equations



v = u + at
s = ut + at^2/2
v^2 = u^2 + 2as


The Attempt at a Solution



I'm pretty sure my first part is correct but here it is anyway:

v = u + at
v = u + (30 + 2t)t
*Sub in values and i get t = 8.508s, then i get s = 1701.652m from the other equation

For the next part I've done this:

a = -0.003v^2
v dv/dx = -0.003v^2
dv/v = -0.003dx *integrate both sides

ln(v) = -0.003x + C

when v = 400, x = 0, C = ln(400)

ln(v) = -0.003x + ln(400)

so when v = 100, i get x = 462.098m

So the total distance i get is 2163.750m but the answer says 2300m...


Where did i go wrong or is the answer given incorrect? :S
 
Physics news on Phys.org
The equations you give only hold for constant acceleration. The acceleration function you give (a=30+2t) is not constant, and so you will need to use calculus to solve even the first part.
 
Ahh whoops, i see it now :)

Now i get the first part with t = 10s and a displacement of 1833.333m

Thanks cristo!
 
eluu said:
Ahh whoops, i see it now :)

Now i get the first part with t = 10s and a displacement of 1833.333m

Thanks cristo!

You're welcome!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
2K