How Do You Solve for Velocity and Position in Calculus Problems?

  • Thread starter Thread starter black_hole
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on solving for velocity and position in calculus problems using the acceleration function a(t) = 1.2t. The correct approach involves finding the velocity function v(t) through integration, resulting in v(t) = 0.6t² + v0, where v0 is determined using the initial condition v(1) = 5 m/s. Additionally, the position function x(t) can be derived similarly by integrating the velocity function. The importance of including an arbitrary constant during integration is emphasized for accurate results.

PREREQUISITES
  • Understanding of calculus concepts, specifically integration and differentiation.
  • Familiarity with the relationship between acceleration, velocity, and position.
  • Knowledge of initial conditions in solving differential equations.
  • Ability to manipulate algebraic expressions and solve for constants.
NEXT STEPS
  • Study the process of finding anti-derivatives in calculus.
  • Learn how to apply initial conditions to solve for constants in integration.
  • Explore the relationship between acceleration, velocity, and position in physics.
  • Practice solving similar calculus problems involving motion and integration.
USEFUL FOR

Students studying calculus, physics enthusiasts, and anyone looking to strengthen their understanding of motion equations in calculus.

black_hole
Messages
71
Reaction score
0

Homework Statement



a9t) = 1.2t. If v(1) = 5m/s, v(2) = ?
If x(1) = 6m, x(2) = ?

Homework Equations





The Attempt at a Solution



I know (anti-deriv) that is something like v(t) = .6t^2, but plug in 1 and that's not 5m/s?
 
Physics news on Phys.org


black_hole said:

Homework Statement



a9t) = 1.2t. If v(1) = 5m/s, v(2) = ?
If x(1) = 6m, x(2) = ?

Homework Equations


The Attempt at a Solution



I know (anti-deriv) that is something like v(t) = .6t^2, but plug in 1 and that's not 5m/s?

It took me the longest time to figure out what you meant by "a9t) = 1.2t"

Do you mean, a(t) = (1.2 [m/s3])t
?

When evaluating indefinite integrals (i.e. anti-derivatives), ensure you add an arbitrary constant to the results.

In other words, for this problem,

v(t) = (0.6 [m/s3])t2 + v0

where v0 is an arbitrary constant. Use v(1 ) = 5 [m/s] to solve for v0.

'Same idea applies when solving for x(t).
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
23
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
902
Replies
4
Views
2K
Replies
16
Views
2K
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K