Question

The equation of motion of a particle is s = t^4 − 3t^3 + t^2 − t, where s is in meters and t is in seconds. (Assume t ≥ 0.)

Find the velocity, v(t), and acceleration, a(t), as functions of t.

Answer 1

The velocity of the particle is v(t) = 4t^3 - 9t^2 + 2t - 1 and the acceleration is a(t) = 12t^2 - 18t + 2, found by differentiating the position function s = t^4 - 3t^3 + t^2 - t first for velocity and then again for acceleration.

Step-by-step explanation:

To find the velocity and acceleration as functions of time for the particle with equation of motion s = t^4 − 3t^3 + t^2 − t, we need to differentiate the position function with respect to time. The velocity, v(t), is the first derivative of the position function, and the acceleration, a(t), is the second derivative of the position function.

The velocity is found by differentiating the position function with respect to t:

v(t) = ds/dt = 4t^3 − 9t^2 + 2t − 1

The acceleration is the derivative of the velocity function:

a(t) = dv/dt = d^2s/dt^2 = 12t^2 − 18t + 2

Therefore, the velocity function of the particle is v(t) = 4t^3 − 9t^2 + 2t − 1 and the acceleration function is a(t) = 12t^2 − 18t + 2.