A particle moves with a fixed acceleration of 3 m/s^2. If its initial velocity is v(0)= 2 m/s, and its initial displacement is s(0)= 0
A)Find its velocity function v(t)
B)How long does it take for the particle to reach the velocity 14 m/s?
C)Find the distance traveled when its velocity reaches 14 m/s.
v = 2 + 3t
when does v=14?
now take the integral to get the distance
Thank you so just to clarify for a) all you did was take the integral of the acceleration to get your v(t) function.
Then for b) do you set v=2+3t equal to 14?
c)You just take the integral of v(t) to get position function I think.
Yes.
The integral of the acceleration vector is the velocity vector. The constant of integration is determined by a known velocity, often the initial velocity.
The integral of velocity is displacement. Constant of integration is determined by initial position (zero here) to give position as a function of time.
A) To find the velocity function v(t), we can use the kinematic equation:
v(t) = v(0) + at
where v(t) is the velocity at time t, v(0) is the initial velocity, a is the constant acceleration, and t is the time elapsed.
Given that the initial velocity v(0) is 2 m/s and the acceleration a is 3 m/s^2, we can substitute these values into the equation:
v(t) = 2 + 3t
Therefore, the velocity function v(t) is 2 + 3t.
B) To find the time it takes for the particle to reach a velocity of 14 m/s, we can use the velocity function v(t) we derived in part A.
We set v(t) = 14 and solve for t:
14 = 2 + 3t
12 = 3t
t = 12/3 = 4 seconds
Therefore, it takes 4 seconds for the particle to reach a velocity of 14 m/s.
C) To find the distance traveled when the velocity reaches 14 m/s, we need to find the displacement function s(t) first.
The displacement function can be obtained by integrating the velocity function:
s(t) = ∫v(t) dt
s(t) = ∫(2 + 3t) dt
s(t) = 2t + (3/2)t^2 + C
where C is the constant of integration.
Since the initial displacement s(0) is 0, we can substitute this value into the displacement function:
0 = 2(0) + (3/2)(0)^2 + C
0 = C
Therefore, C = 0 and the final displacement function is:
s(t) = 2t + (3/2)t^2
To find the distance traveled when the velocity reaches 14 m/s, we need to find the displacement at that time.
Substituting t = 4 into the displacement function:
s(4) = 2(4) + (3/2)(4)^2
s(4) = 8 + (3/2)(16)
s(4) = 8 + 24
s(4) = 32 meters
Therefore, the distance traveled when the velocity reaches 14 m/s is 32 meters.