At t=0 , a particle starts at the origin with a velocity of 6 feet per second and moves along the x-axis in such a way that at time t its acceleration is 12t^2 feet per second per second. Through how many feet does the particle move during the first 2 seconds?
a = 12t^2
velocity = 4t^3 + c
when t = 0, v = 6
6 = 4(0) + c
c = 6
v = 4t^3 + 6
distance = t^4 + 6t + k
when t = 0, distance = 0 ---> k = 0
distance = t^4 + 6t
so when t = 2
distance = 2^4 + 6(2) = 28 feet
Thanks, Reiny!
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Because it realized it could move feet and still get a lot of laughs!
To calculate how many feet the particle moves during the first 2 seconds, we can integrate the acceleration function with respect to time twice, and then evaluate the definite integral from 0 to 2.
The integral of the acceleration function, 12t^2, with respect to time gives us the velocity function. Integrating again gives us the position function.
Integrating once, we have:
∫(12t^2) dt = 4t^3
So the velocity function is: v(t) = 4t^3 + C
To find the constant C, we use the initial condition that at t = 0, the velocity is 6 ft/s:
v(0) = 4(0^3) + C = 6
C = 6
So the velocity function is: v(t) = 4t^3 + 6
Integrating again, we have:
∫(4t^3+6) dt = t^4 + 6t + D
To find D, we use the initial condition that at t = 0, the particle is at the origin:
x(0) = 0^4 + 6(0) + D = 0
D = 0
So the position function is: x(t) = t^4 + 6t
To find how many feet the particle moves during the first 2 seconds, we evaluate the position function from 0 to 2:
x(2) - x(0) = (2^4 + 6(2)) - (0^4 + 6(0)) = 32 + 12 - 0 = 44 feet
Therefore, the particle moves 44 feet during the first 2 seconds.
To find the distance moved by the particle during the first 2 seconds, we need to determine the particle's position function by integrating its velocity function.
Given:
Initial velocity (v0) = 6 feet per second
Acceleration (a) = 12t^2 feet per second per second
We know that the velocity function can be obtained by integrating the acceleration function with respect to time:
v(t) = ∫(12t^2) dt
v(t) = 12 ∫(t^2) dt
Integrating t^2 with respect to t gives:
v(t) = 12 * (t^3 / 3) + C
v(t) = 4t^3 + C
Since the particle starts at the origin at t=0 with an initial velocity of 6 feet per second, we can substitute these initial conditions to find the constant C:
v(0) = 4(0)^3 + C
6 = 0 + C
C = 6
Now we have the velocity function:
v(t) = 4t^3 + 6
To find the position function, we need to integrate the velocity function with respect to time:
s(t) = ∫(4t^3 + 6) dt
s(t) = 4 ∫(t^3) dt + 6 ∫(1) dt
Integrating t^3 with respect to t gives:
s(t) = 4 * (t^4 / 4) + 6 * t + C
s(t) = t^4 + 6t + C
Since the particle starts at the origin, we can substitute t=0 to find the constant C:
s(0) = (0)^4 + 6(0) + C
0 = 0 + 0 + C
C = 0
Now we have the position function:
s(t) = t^4 + 6t
To find the distance moved during the first 2 seconds, we can subtract the initial position (s(0) = 0) from the position at t=2:
Distance moved = s(2) - s(0)
Distance moved = (2^4 + 6(2)) - 0
Distance moved = (16 + 12) - 0
Distance moved = 28 feet
Therefore, the particle moves 28 feet during the first 2 seconds.
To find the total distance the particle moves during the first 2 seconds, we need to determine the particle's position at time t = 2 and subtract its initial position at t = 0.
To do this, we will integrate the velocity function to obtain the position function, then evaluate it at t = 2.
Given:
Velocity function (v): 6 feet per second
Acceleration function (a): 12t^2 feet per second per second
Step 1: Integrate the acceleration function
To find the velocity function, we integrate the acceleration function with respect to time (t):
v = ∫(a)dt = ∫(12t^2)dt
Integrating 12t^2 with respect to t gives us:
v = 4t^3 + C
Since the particle starts with a velocity of 6 feet per second, we substitute t = 0 and v = 6 into the velocity function:
6 = 4(0)^3 + C
6 = 0 + C
C = 6
Therefore, the velocity function is:
v = 4t^3 + 6
Step 2: Find the position function
To find the position function, we integrate the velocity function with respect to time (t):
s = ∫(v)dt = ∫(4t^3 + 6)dt
Integrating 4t^3 with respect to t gives us:
s = t^4 + 6t + D
Step 3: Evaluate the position at t = 2
To find the particle's position at t = 2, substitute t = 2 into the position function:
s(2) = (2)^4 + 6(2) + D
Simplifying this expression gives us:
s(2) = 16 + 12 + D
s(2) = 28 + D
Step 4: Determine the initial position
The particle starts at the origin, which means its initial position at t = 0 is s(0) = 0.
Substituting t = 0 into the position function:
s(0) = (0)^4 + 6(0) + D
0 = 0 + 0 + D
D = 0
Therefore, the position function is:
s = t^4 + 6t
Step 5: Calculate the distance traveled in the first 2 seconds
To find the total distance traveled in the first 2 seconds, subtract the initial position from the position at t = 2:
Distance = |s(2) - s(0)| = |28 + 0 - 0|
Distance = |28|
Therefore, the particle moves 28 feet during the first 2 seconds.