Question

What is the derivative with respect to time dx/dt of the bowling ball's position-time relationship (x(t) = v0t + x0, where v0 = 2.5 m/s and x0 = -5.0 m)?

Answer 1

The derivative of the bowling ball's position-time relationship, dx/dt, represents its velocity. In the given equation x(t) = v0t + x0 with v0 = 2.5 m/s and x0 = -5.0 m, the derivative with respect to time, dx/dt, is the constant velocity v0, which is 2.5 m/s.

Step-by-step explanation:

The derivative of the bowling ball's position-time relationship with respect to time, dx/dt, is the velocity of the bowling ball. Given the position-time relationship x(t) = v0t + x0, where v0 = 2.5 m/s and x0 = -5.0 m, the derivative is simply the coefficient of t in the equation, which is the initial velocity, v0. Therefore, when differentiating x(t) with respect to t, dx/dt equals 2.5 m/s, indicating the bowling ball travels at a constant velocity of 2.5 m/s.