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.
Hey Stan, is that a direct copy/paste? Ie that's all your were given word for word? Hoi.. I think my Physics is rusty.
yeah that is word for word
Hey again Stan, hmm, okay, I'll keep playing with it, I still say we're missing info though... but re-post in the meantime, so other tutors don't think I've answered.
Acceleration is constant. Initial velocity is zero. You know that from the first sentence.
Vertical location is
Y = -2 + (a/2)t^2
Y(t=2) - Y(t=1) = 3 ft.
(a/2) (2^2 - 1^2)) = (3/2)a = 3
a = 2 ft/s^2
Y(t=3) = -2 + (2/2)*3^2 = 7 feet
To solve this problem, we can use the given information about the velocity and displacement of the particle.
Let's first analyze the information given:
1. The particle starts from rest 2 ft below the origin when t = 0.
2. The magnitude of the velocity of the particle is directly proportional to the time after starting.
3. The displacement of the particle during the time interval from t = 1 sec to t = 2 sec is 3 ft upward.
Let's break down the problem into steps:
Step 1: Determine the velocity of the particle
Since the magnitude of the velocity is directly proportional to the time, we can write this relationship as:
v = kt
where v is the velocity and k is the constant of proportionality.
Since the particle starts from rest at t = 0, we can substitute t = 0 into the equation to find the value of k:
0 = k(0)
0 = 0
This means that k = 0.
Therefore, the equation for the velocity becomes:
v = 0t
v = 0 ft/sec
Step 2: Determine the displacement of the particle
To find the displacement of the particle, we can integrate the velocity function with respect to time.
∫v dt = ∫0 dt = 0t + C
s = Ct + D
where s is the displacement and C and D are constants of integration.
Given the displacement during the time interval from t = 1 sec to t = 2 sec is 3 ft upward, we can substitute these values into the equation:
3 = C(2) + D ---(1)
Substituting t = 0 into the equation gives us the initial position:
-2 = C(0) + D
-2 = D ---(2)
Substituting equation (2) into equation (1):
3 = C(2) - 2
5 = 2C
C = 5/2
Therefore, the equation for the displacement becomes:
s = (5/2)t - 2
Step 3: Determine the location of the particle when t = 3 sec (a)
To find the location when t = 3 sec, we substitute t = 3 into the equation for displacement:
s = (5/2)(3) - 2
s = 15/2 - 2
s = 11/2 or 5.5 ft
Therefore, when t = 3 sec, the location of the particle is 5.5 ft.
Step 4: Determine the acceleration of the particle (b)
Acceleration is the derivative of velocity with respect to time.
a = dv/dt
Since the velocity equation is v = 0t, the derivative of this equation with respect to time is:
a = d(0t)/dt
a = 0
Therefore, the acceleration of the particle is 0 ft/sec².
In summary:
(a) When t = 3 sec, the location of the particle is 5.5 ft.
(b) The acceleration of the particle is 0 ft/sec².