Question

Work through the mathematics of how a constant force acting on an object, in one dimension, will affect the velocity as a function of time and the position as a function of time (definite integrals will be involved).  Then combine the results for position and velocity to derive the relationship between position and speed for an object that moves in one dimension with a constant force (hint:  algebraically eliminate time).  What situation(s) might this theory describe?

Answer 1

Answer 2

Let's consider an object with mass "m" moving in one dimension under the influence of a constant force "F." We'll work through the mathematics step-by-step to derive the relationship between position and speed.

1. Newton's Second Law of Motion: Newton's second law states that the acceleration "a" of an object is directly proportional to the force acting on it and inversely proportional to its mass:

a = F / m

2. Velocity as a function of time: The acceleration is the rate of change of velocity with respect to time. We can express this as a differential equation:

dv/dt = a

Where "v" is the velocity of the object.

3. Solving the differential equation: To find the velocity as a function of time, we integrate both sides with respect to time:

∫dv = ∫a dt

∫dv = ∫(F / m) dt

Integrating gives:

v = (F / m) * t + C1

Where "C1" is the constant of integration, which represents the initial velocity of the object.

4. Position as a function of time: The velocity is the rate of change of position with respect to time. We can express this as another differential equation:

dx/dt = v

Where "x" is the position of the object.

5. Solving the differential equation: To find the position as a function of time, we integrate both sides with respect to time:

∫dx = ∫v dt

∫dx = ∫[(F / m) * t + C1] dt

Integrating gives:

x = (F / (2 * m)) * t^2 + C1 * t + C2

Where "C2" is the constant of integration, which represents the initial position of the object.

6. Relationship between position and speed: Now, we can combine the results for position and velocity to eliminate time. We have two equations:

v = (F / m) * t + C1 x = (F / (2 * m)) * t^2 + C1 * t + C2

Let's solve the first equation for "t":

t = (v - C1) * (m / F)

Now, substitute this value of "t" into the second equation:

x = (F / (2 * m)) * [(v - C1) * (m / F)]^2 + C1 * [(v - C1) * (m / F)] + C2

Simplify:

x = (m / (2 * F)) * (v - C1)^2 + C1 * (v - C1) * (m / F) + C2

x = (m / (2 * F)) * (v - C1)^2 + (m / F) * C1 * (v - C1) + C2

x = (m / (2 * F)) * [v^2 - 2 * v * C1 + C1^2] + (m / F) * [C1 * v - C1^2] + C2

Combine like terms:

x = (m / (2 * F)) * v^2 + (m / F) * C1 * v + (m / (2 * F)) * C1^2 - (m / F) * C1^2 + C2

x = (m / (2 * F)) * v^2 + [(m / F) * C1 - (m / F) * C1^2] * v + [(m / (2 * F)) * C1^2 + C2]

Now, let's define some constants:

k1 = m / (2 * F) k2 = m / F k3 = (m / (2 * F)) * C1^2 + C2

The position-speed relationship can now be expressed as:

x = k1 * v^2 + k2 * v + k3

This equation represents a quadratic relationship between position (x) and speed (v) for an object moving in one dimension under the influence of a constant force. This situation describes the motion of an object undergoing constant acceleration, like a projectile moving vertically under the influence of gravity, or an object subjected to a constant force along a straight path (e.g., a car moving on a level road with a constant engine force).