If the position of an object is given by x = 2.00t^3, where x is measured in meters and t in seconds, find (a) the average velocity and (b) the average acceleration between t = 1.0 s and t = 2.0 s. Then find (c) the instantaneous velocity v and (d) the instantaneous acceleration a at t = 1.0 s. Next find (e)v and (f)a at t = 2.0 s.
x = 2t^3
v = 6t
a = 6
To find an average, just divide the change by the interval.
(a)
x(1) = 2
x(2) = 16
avg v = ∆x/∆t = (16-2)m/(2-1)s = 14m/s
I think you can handle the rest, no?
No
To find the average velocity between t = 1.0 s and t = 2.0 s, we need to find the change in position (Δx) divided by the change in time (Δt) during that interval.
(a) Average Velocity:
Δx = x2 - x1 = (2.00t2^3)|t=2 - (2.00t1^3)|t=1 = 2.00(2^3) - 2.00(1^3) = 14.00 m^3 - 2.00 m^3 = 12.00 m^3
Δt = t2 - t1 = 2.00 s - 1.00 s = 1.00 s
Average Velocity (v_avg) = Δx/Δt = 12.00 m^3 / 1.00 s = 12.00 m/s
Next, to find the average acceleration between t = 1.0 s and t = 2.0 s, we need to find the change in velocity (Δv) divided by the change in time (Δt) during that interval.
(b) Average Acceleration:
Δv = v2 - v1
Δv = (v_avg)t2 - (v_avg)t1
Δv = (v_avg)(2.00 s) - (v_avg)(1.00 s)
Δv = 12.00 m/s * 2.00 s - 12.00 m/s * 1.00 s
Δv = 24.00 m - 12.00 m = 12.00 m/s
Δt = t2 - t1 = 2.00 s - 1.00 s = 1.00 s
Average Acceleration (a_avg) = Δv/Δt = 12.00 m/s / 1.00 s = 12.00 m/s^2
To find the instantaneous velocity (v) at t = 1.0 s, we can take the derivative of the given position function x = 2.00t^3 with respect to time (t).
(c) Instantaneous Velocity at t = 1.0 s:
v = dx/dt
v = d(2.00t^3)/dt
v = 2.00 * d(t^3)/dt
v = 2.00 * 3t^2
v = 6.00t^2
Substituting t = 1.0 s into the equation:
v = 6.00(1.0 s)^2
v = 6.00 m/s^2
To find the instantaneous acceleration (a) at t = 1.0 s, we can take the derivative of the instantaneous velocity function v = 6.00t^2 with respect to time (t).
(d) Instantaneous Acceleration at t = 1.0 s:
a = dv/dt
a = d(6.00t^2)/dt
a = 6.00 * d(t^2)/dt
a = 6.00 * 2t
a = 12.00t
Substituting t = 1.0 s into the equation:
a = 12.00(1.0 s)
a = 12.00 m/s^2
Finally, to find the instantaneous velocity (v) and instantaneous acceleration (a) at t = 2.0 s, we can substitute t = 2.0 s into the equations we derived above.
(e) Instantaneous Velocity at t = 2.0 s:
v = 6.00(2.0 s)^2
v = 6.00 * 4.0 s^2
v = 24.00 m/s
(f) Instantaneous Acceleration at t = 2.0 s:
a = 12.00(2.0 s)
a = 24.00 m/s^2
Therefore, the answers are:
(a) Average Velocity = 12.00 m/s
(b) Average Acceleration = 12.00 m/s^2
(c) Instantaneous Velocity at t = 1.0 s = 6.00 m/s
(d) Instantaneous Acceleration at t = 1.0 s = 12.00 m/s^2
(e) Instantaneous Velocity at t = 2.0 s = 24.00 m/s
(f) Instantaneous Acceleration at t = 2.0 s = 24.00 m/s^2