Acceleration function of velocity example

Click For Summary
SUMMARY

The discussion focuses on the acceleration function of a bicyclist coasting down a hill, modeled by the equation dv/dt = a - cv, where 'a' and 'c' are constants. The velocity as a function of distance is derived as x = -v/c - (a/c^2)*ln(a-cv)/a, with the maximum velocity attained when acceleration equals zero. The conversation highlights the relationship between velocity and distance, emphasizing the role of drag proportional to the square of the velocity.

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with calculus concepts, particularly integration
  • Knowledge of kinematics, specifically velocity and acceleration
  • Basic grasp of logarithmic functions
NEXT STEPS
  • Study the derivation of solutions for first-order differential equations
  • Explore the concept of terminal velocity in physics
  • Learn about the impact of drag forces on motion
  • Investigate numerical methods for solving differential equations
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion under the influence of drag forces, particularly in the context of cycling and similar scenarios.

renn12
Messages
2
Reaction score
0

Homework Statement



Because the drag on objects moving through air increases as the square
of the velocity, the acceleration of a bicyclist coasting down a slight hill
is (v) = a - cv where a and c are constant. Determine the velocity of
the bicyclist as a function of distance if the velocity is zero when x=0.
Also determine the maximum velocity that the cyclist attains.

Homework Equations


dx/dt = v
dv/dt = a-cv

The Attempt at a Solution


x = -v/c - (a/c^2)*ln(a-cv)/a
v = (a - ae^-ct)/c
t = -(cx + v)/a = -(1/c)*ln(a-cv)/a
 
Last edited:
Physics news on Phys.org
"...the acceleration of a bicyclist coasting down a slight hill
is (v) = a - cv where a and c are constant."

Shouldn't the v on LHS be dv/dt?

Hint for v as a function of distance:

V = dx/dt = (dx/dv)*(dv/dt) = (dx/dv)*(a - c*V)

Maximum velocity occurs when acceleration is zero. which is what your second equation shows for large t.
 

Similar threads

  • · Replies 69 ·
3
Replies
69
Views
6K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K