A particles position is described by the function x=3.4t^2 - 2.2t + 2.7 (m). Determine the average acceleration (in m/s^2) for the interval of time t1= 1.0s and t2= 3.0s
v = 6.8t - 2.2
avg acceleration is
(v(3)-v(1))/(3-1)
To determine the average acceleration for the interval of time between t1 = 1.0s and t2 = 3.0s, we need to find the velocity function first. The velocity function is the derivative of the position function with respect to time.
Given the position function x = 3.4t^2 - 2.2t + 2.7 (m).
To find the velocity function, we need to differentiate the position function with respect to time (t):
dx/dt = d/dt(3.4t^2 - 2.2t + 2.7)
= 6.8t - 2.2
The velocity function is given by dx/dt = 6.8t - 2.2 (m/s).
To find the average acceleration, we need to find the change in velocity and divide it by the change in time.
Change in velocity (Δv) = v2 - v1, where v2 = velocity at t2 and v1 = velocity at t1.
Substituting t2 = 3.0s and t1 = 1.0s into the velocity function:
v2 = (6.8 * 3.0) - 2.2
= 20.4 - 2.2
= 18.2 m/s
v1 = (6.8 * 1.0) - 2.2
= 6.8 - 2.2
= 4.6 m/s
Change in velocity (Δv) = v2 - v1
= 18.2 - 4.6
= 13.6 m/s
Now, we need to find the change in time (Δt) = t2 - t1 = 3.0 - 1.0 = 2.0s.
Finally, we can calculate the average acceleration using the formula:
Average Acceleration = Δv / Δt
= 13.6 / 2.0
= 6.8 m/s^2
Therefore, the average acceleration for the interval of time between t1 = 1.0s and t2 = 3.0s is 6.8 m/s^2.