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.