We can express the velocity as a function of time, v(t), of a car as v(t) = (8.3 m/s) + (6.1 m/s2)t - (4.2 m/s3)t2. What can you say about the average acceleration of the car between t = 1.0 s and t = 2.0 s?

A.13 m/s2


B.-6.3 m/s2


C.-13 m/s2


D.0 m/s2


E. 15 m/s2

average acceleration=changevelocity/time

= (v(2)-v(1))/(2-1)

what do i plug in for the equation?

v(2)= the velocity at time 2 from your equation. v(1) is velocity at time 1.

If you are looking to learn physics by plugging in numbers to formulas, you are headed for a long year.

To find the average acceleration of the car between t = 1.0 s and t = 2.0 s, we need to find the change in velocity and divide it by the change in time.

The change in velocity (Δv) can be found by calculating the velocity at t = 2.0 s (v(2.0)) minus the velocity at t = 1.0 s (v(1.0)):

Δv = v(2.0) - v(1.0)

Substituting the given equation for velocity, we have:

Δv = [(8.3 m/s) + (6.1 m/s^2)(2.0 s) - (4.2 m/s^3)(2.0 s)^2] - [(8.3 m/s) + (6.1 m/s^2)(1.0 s) - (4.2 m/s^3)(1.0 s)^2]

Simplifying the equation, we get:

Δv = [8.3 m/s + 12.2 m/s - 16.8 m/s] - [8.3 m/s + 6.1 m/s - 4.2 m/s]

Δv = 3.7 m/s - 10.2 m/s

Δv = -6.5 m/s

The change in time (Δt) is 2.0 s - 1.0 s = 1.0 s.

Now, we can calculate the average acceleration using the formula:

Average Acceleration = Δv / Δt

Average Acceleration = (-6.5 m/s) / (1.0 s)

Average Acceleration = -6.5 m/s^2

Therefore, the average acceleration of the car between t = 1.0 s and t = 2.0 s is approximately -6.3 m/s^2.

Hence, the correct answer is B. -6.3 m/s^2.