Prove that if a particle moves along a curve at constant speed, then the velocity vector is at all times perpendicular to the acceleration vector.
To prove that if a particle moves along a curve at constant speed, then the velocity vector is at all times perpendicular to the acceleration vector, we can use a mathematical approach. Here are the steps to prove this statement:
Step 1: Define the terms:
- Particle: The object that is moving along the curve.
- Curve: The path along which the particle is moving.
- Speed: The rate at which the particle covers the distance along the curve.
- Velocity: The rate of change of the particle's position with respect to time.
- Acceleration: The rate of change of the particle's velocity with respect to time.
Step 2: Use parametric equations:
Let's represent the curve using parametric equations, where the position of the particle at any given time t is described as (x(t), y(t)).
Step 3: Derive the velocity vector:
To find the velocity vector, differentiate the position vector with respect to time:
Velocity vector V(t) = (x'(t), y'(t))
Step 4: Derive the acceleration vector:
To find the acceleration vector, differentiate the velocity vector with respect to time:
Acceleration vector A(t) = (x''(t), y''(t))
Step 5: Prove the statement:
Since the particle is moving at constant speed, the magnitude of the velocity vector V(t) remains constant:
|V(t)| = √(x'(t)^2 + y'(t)^2) = constant
Now, we can prove that the velocity vector is perpendicular to the acceleration vector A(t) by showing that their dot product is zero at all times:
V(t) · A(t) = (x'(t), y'(t)) · (x''(t), y''(t))
= x'(t) * x''(t) + y'(t) * y''(t)
Since the speed is constant, |V(t)| is constant, and its derivative with respect to time is zero:
d(|V(t)|)/dt = 0
Differentiating |V(t)|^2 = (x'(t)^2 + y'(t)^2) with respect to time yields:
2 * V(t) · A(t) = 2 * x'(t) * x''(t) + 2 * y'(t) * y''(t)
Now, substitute the fact that d(|V(t)|)/dt = 0 into the equation:
0 = 2 * x'(t) * x''(t) + 2 * y'(t) * y''(t)
0 = x'(t) * x''(t) + y'(t) * y''(t)
Since the right-hand side of the equation is the dot product of the velocity vector and acceleration vector, we can conclude that V(t) and A(t) are perpendicular at all times, which proves the statement.
Therefore, if a particle moves along a curve at constant speed, the velocity vector is always perpendicular to the acceleration vector.