Question

The maximum acceleration attained on the interval 0≤t≤3 by the particle whose velocity is given by v(t) = t3 -3t2 +12t +4 is? a.9

b.12
c.14
d.21
e.40

Answer 1

The maximum acceleration attained on the interval
`\(0 \leq t \leq 3\)` by the particle whose velocity is given by
`\(v(t) = t^3 - 3t^2 + 12t + 4\)`is 21 (Option d).Thus the correct option is:d.21

Step-by-step explanation:

To find the acceleration, we need to take the derivative of the velocity function with respect to time, \(t\). The velocity function is
`\(v(t) = t^3 - 3t^2 + 12t + 4\).` Taking the derivative, we get the acceleration function
`\(a(t)\).`

`\[a(t) = v'(t) = 3t^2 - 6t + 12\]`

Now, we want to find the maximum acceleration on the interval
`\(0 \leq t`
`\leq 3\).` To find critical points, set \
`(a'(t) = 0\).`

`\[a'(t) = 6t - 6\]`

Setting
`\(6t - 6 = 0\)`and solving for
`\(t\),`we find
`\(t = 1\)`. Now, we check the endpoints of the interval and the critical point by evaluating
`\(a(t)\) at \(t = 0, 1, \text{ and } 3\).`

`\[a(0) = 12, \quad a(1) = 9, \quad a(3) = 21\]`

Comparing the values, we see that the maximum acceleration occurs at
`\(t = 3\)`with \
`(a(3) = 21\).` Therefore, the correct answer is 21, corresponding to option d. The particle attains its maximum acceleration of 21 on the given interval.Thus the correct option is:d.21

Answer 2

The maximum acceleration attained on the interval 0≤t≤3 for the particle with velocity function v(t) = t³ - 3t² + 12t + 4 is 12 m/s².

Step-by-step explanation:

To find the maximum acceleration attained on the interval 0≤t≤3 for the particle with velocity function v(t) = t³ - 3t² + 12t + 4, we need to find the maximum of the acceleration function a(t) by taking its derivative.

To do this, we calculate the derivative of v(t) with respect to t, which is a(t).

a(t) = dv(t)/dt = d/dt(t³ - 3t² + 12t + 4)

By differentiating term-by-term, we get:

a(t) = 3t² - 6t + 12

To find the maximum acceleration, we set a(t) equal to zero and solve for t:

3t² - 6t + 12 = 0

Using the quadratic formula, we find two roots: t = 2 and t = -2.

Since the interval for t is 0≤t≤3, we only consider the positive root, t = 2.

Therefore, the maximum acceleration attained on the interval 0≤t≤3 is a(t) = 3(2)² - 6(2) + 12 = 12 m/s².