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:
v(t) = 2t³-t²-4
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:
v(2) — v(0) (2) - (0) avg =
(8) - (-4) m/s
2s
= 6 m/s²
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:
a(t) = d dt [v(t)]
Substitute and solve:
a(t)=
-
d
dt
= 6t²2t
To find the time(s) for which the acceleration is equal to 4, set a equal to 4 and solve for t:
4 = 6t² - 2t
2 = 3t²-t
3t²t-20
Mathematics
10 points
(t-1)(3t+ 2) = 0
2 3 t = 1 or t =
Since t≥ 0, we can ignore the second solution.
Evaluate v at t = 1:
v(t) = 2(1)³ (1)² - 4 = −3 m/s
In conclusion, when the acceleration is 4 m/s² at t= 1, the velocity is -3 m/s.