Engineering Dynamics (Motion related)

Click For Summary
SUMMARY

The discussion centers on a physics problem involving a rocket sled that accelerates and then decelerates. The sled accelerates at a rate of a = 30 + 2t until reaching a velocity of 400 m/s, followed by a deceleration described by a = -0.003v^2 until its velocity decreases to 100 m/s. The participant calculated the total distance traveled as 2163.750 meters, while the expected answer is 2300 meters. The discrepancy suggests a potential error in the integration or assumptions made during the calculations.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of motion dynamics, including acceleration and deceleration
  • Experience with solving differential equations related to motion
NEXT STEPS
  • Review the derivation of kinematic equations for uniformly accelerated motion
  • Study integration methods for solving differential equations in motion problems
  • Learn about the implications of variable acceleration in dynamics
  • Examine case studies involving real-world applications of motion dynamics
USEFUL FOR

Students and professionals in physics, engineering, and applied mathematics who are tackling problems related to motion dynamics and require a deeper understanding of acceleration and deceleration principles.

eluu
Messages
9
Reaction score
0

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

Similar threads

  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K