Question

An object moves along the x-axis so that at any time t ≥ 0, its velocity is given by

v(t) = 4 − 3.8 cos(0.7t). What is the acceleration of the particle at time t = 6?

A) -0.913
B) 1.098
C) -2.318
D) 5.863

Answer 1

The acceleration of the particle at time t = 6 is found by taking the derivative of the velocity function, applying the chain rule, and substituting t = 6 into the resulting acceleration function.

Step-by-step explanation:

To find the acceleration of the particle at a specific time t, we need to take the derivative of the velocity function v(t). Given that v(t) = 4 − 3.8 cos(0.7t), we differentiate this function with respect to time t to get the acceleration function a(t). The derivative of the cosine function is − sine, and we also need to apply the chain rule to take into account the constant 0.7 that is multiplied by t.

The derivative is a(t) = d/dt [4 − 3.8 cos(0.7t)] = 0 + 3.8 × 0.7 × sin(0.7t), which simplifies to a(t) = 2.66 sin(0.7t). Plugging in t = 6, we find a(6) = 2.66 sin(0.7 × 6).

By calculating this value, we can then determine which of the given options (A, B, C, or D) represents the acceleration of the particle at t = 6 seconds.