Answer:
Part A)
6 m/s²
Part B)
-3 m/s.
Step-by-step explanation:
We are given that the velocity, in meters per second, of an object at a time t for t ≥ 0 is given by the function:
[tex]\displaystyle v(t) = 2t^3 - t^2 - 4[/tex]
Part A)
To find the average acceleration of the object during the first two seconds, we can find the average slope of v from t = 0 to t = 2. Hence:
[tex]\displaystyle \begin{aligned} \text{avg} & = \frac{v(2) - v(0)}{(2)-(0)} \\ \\ & = \frac{(8) - (-4)\text{ m/s}}{2\text{ s}}\\ \\ & = 6\text{ m/s$^2$}\end{aligned}[/tex]
Hence, the average acceleration of the object during the first two seconds was 6 m/s².
Part B)
Recall that acceleration is the derivative of the velocity function. In other words:
[tex]\displaystyle a(t) = \frac{d}{dt}\left[ v(t)\right][/tex]
Substitute and solve:
[tex]\displaystyle \begin{aligned}a(t) & = \frac{d}{dt}\left[ 2t^3 - t^2 - 4\right] \\ \\ &= 6t^2 - 2t \end{aligned}[/tex]
To find the time(s) for which the acceleration is equal to 4, set a equal to 4 and solve for t:
[tex]\displaystyle \begin{aligned}4 & = 6t^2 - 2t \\ \\ 2 & = 3t^2 - t \\ \\ 3t^2 - t - 2 & = 0 \\ \\ (t-1)(3t+2) & = 0 \\ \\ t = 1 \text{ or } t = -\frac{2}{3} \end{aligned}[/tex]
Since t ≥ 0, we can ignore the second solution.
Evaluate v at t = 1:
[tex]\displaystyle v(t) = 2(1)^3 - (1)^2 - 4 = -3\text{ m/s}[/tex]
In conclusion, when the acceleration is 4 m/s² at t = 1, the velocity is -3 m/s.
