The magnitude of the velocity of a particle
which starts from rest 2 ft below the origin when
t = 0 and moves along a vertical axis is directly
proportional to the time after starting. The
displacement of the particle during the time
interval from t = 1 sec to t = 2 sec is 3 ft upward.
When t = 3 sec, determine (a) the location of the
particle; (b) the acceleration of the particle.
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To determine the location of the particle at t = 3 sec, we need to integrate the velocity function over the interval from t = 0 to t = 3 sec.
Given that the magnitude of the velocity of the particle is directly proportional to the time after starting, we can express the velocity function as v(t) = kt, where k is a constant.
To find the value of k, we can use the given information that the displacement of the particle during the time interval from t = 1 sec to t = 2 sec is 3 ft upward.
We know that displacement is the integral of velocity. So, integrating v(t) from t = 1 sec to t = 2 sec, we get:
∫[v(t)]dt = ∫[kt]dt = ½k[t^2] from 1 to 2.
Since the displacement is 3 ft upward, we can set up the equation:
3 = ½k(2^2 - 1^2) = ½k(4 - 1).
Simplifying the equation, we get:
3 = ½k(3).
Now, solve for k:
k = 2.
Therefore, the velocity function is v(t) = 2t.
To determine the location of the particle at t = 3 sec, we integrate the velocity function over the interval from t = 0 to t = 3 sec:
∫[2t]dt = t^2 + C,
where C is a constant of integration.
At t = 0, the particle starts 2 ft below the origin. Therefore, at t = 0, the location of the particle is:
0^2 + C = C.
So, the location of the particle at t = 0 is C.
Now, at t = 3 sec, the location of the particle is:
(3^2 + C) ft above the origin.
Therefore, the location of the particle at t = 3 sec is 9 + C ft above the origin.
To find the value of C, we can use the given information that the particle is 2 ft below the origin at t = 0. So, C = -2.
Therefore, the location of the particle at t = 3 sec is (9 -2) ft above the origin, which is 7 ft.
So, the location of the particle at t = 3 sec is 7 ft above the origin.
To determine the acceleration of the particle, we take the derivative of the velocity function with respect to time:
a(t) = dv(t)/dt = d(2t)/dt = 2.
Therefore, the acceleration of the particle is a(t) = 2 ft/sec^2.