Solving Kinematics Problem: Integrate for x,v,a @ t=3s

Click For Summary
SUMMARY

The discussion focuses on solving a kinematics problem involving a particle with acceleration defined as a = 2√v, where v is the velocity expressed as v = dx/dt. At t = 2s, the particle's position is x = (64/3)m and its velocity is v = 16m/s. The solution involves integrating the differential equation dv/dt = 2√v, leading to the conclusion that by squaring the integrated expression and applying initial conditions, one can determine the particle's location and acceleration at t = 3s.

PREREQUISITES
  • Understanding of basic calculus and integration techniques
  • Familiarity with kinematics concepts such as velocity and acceleration
  • Knowledge of differential equations and their solutions
  • Ability to manipulate algebraic expressions involving square roots
NEXT STEPS
  • Study the method of integrating differential equations, particularly those involving square roots
  • Learn about initial value problems in kinematics and how to apply them
  • Explore advanced integration techniques, including substitution and integration by parts
  • Investigate the physical interpretation of acceleration and velocity in motion problems
USEFUL FOR

Students and professionals in physics, engineering, and mathematics who are tackling problems related to motion and kinematics, particularly those involving differential equations.

naggy
Messages
58
Reaction score
0
A particle moves along a straight line with accel. a = 2\sqrt{v} where v = \frac{dx}{dt} is the velocity. At t = 2s, the particle is located at x= (64/3)m with velocity v = 16m/s. Find the location and acceleration at time t = 3s (hint: integrate)

Now, if the differential equation is dv/dt = 2\sqrt{v}. I don´t know how to solve such an equatoin (because of the square root). But then I thought v must be constant for the integration to work, but that would mean that the acceleration would be zero and that makes no sense. I need help.
 
Physics news on Phys.org
never mind, I figured it out as soon as I clicked submit post.

dv over the sqrt(v) = 2dt

No w8, this makes no sense, if you integrate you get 2sqrt(v) = 2t + c
 
Last edited:
Square both sides of that expression and integrate again. You have enough data to find the constants.
 

Similar threads

Confused about kinematic equations
  • · Replies 20 ·
Replies
20
Views
3K
Particle Motion; Graphical Method
  • · Replies 9 ·
Replies
9
Views
2K
Pushing a stalled car out of an intersection
  • · Replies 1 ·
Replies
1
Views
2K
Kinematics problem with differential equations.
  • · Replies 4 ·
Replies
4
Views
2K
Solving for Distance, Velocity and Acceleration
  • · Replies 3 ·
Replies
3
Views
2K
Having a problem in steps while solving integrals
  • · Replies 6 ·
Replies
6
Views
2K
The gravitating of a small mass towards a big mass
  • · Replies 3 ·
Replies
3
Views
2K
Finding Initial Velocity v0x for Particle with Non-Constant Acceleration
  • · Replies 2 ·
Replies
2
Views
4K
Non-constant acceleration, solving for velocity
  • · Replies 11 ·
Replies
11
Views
3K
Friction and kinematics problem
  • · Replies 7 ·
Replies
7
Views
2K