A particle is moving with the given data. Find the position of the particle.
a(t) = t^2 − 8t + 5, s(0) = 0, s(1) = 20
are you using the usual notation
so that a(t) is acceleration, v(t) is velocity and s(t) is distance ??
I will assume you are
a(t) = t^2 - 8t + 5
v(t) = (1/3)t^3 - 4t^2 + 5t + c
s(t) = (1/12)t^4 - (4/3)t^3 + (5/2)t^2 + ct + k
s(0) = 0
0-0+0+0+k=0
k=0
s(1) = 20
1/12 - 4/3 + 5/2 + c = 0 = 20
5/4 + c = 20
c = 75/4
s(t) = (1/12)t^4 - (4/3)t^3 + (5/2)t^2 + (75/4)t
To find the position of the particle, we need to integrate the given function a(t) to find the velocity function v(t), and then integrate v(t) to find the position function s(t).
Given:
a(t) = t^2 - 8t + 5
s(0) = 0
s(1) = 20
Step 1: Integrate a(t) to find v(t)
To integrate a(t), we need to find the antiderivative of t^2 - 8t + 5.
∫(t^2 - 8t + 5) dt = (1/3)t^3 - 4t^2 + 5t + C
Let v(t) = (1/3)t^3 - 4t^2 + 5t + C, where C is the constant of integration.
Step 2: Find the value of C
To find the constant of integration C, we can use the given condition s(0) = 0.
s(t) = ∫v(t) dt = ∫[(1/3)t^3 - 4t^2 + 5t + C] dt
= (1/12)t^4 - (4/3)t^3 + (5/2)t^2 + Ct + D
Using s(0) = 0, we can substitute t = 0 and s(t) = 0 to find D.
0 = (1/12)*(0)^4 - (4/3)*(0)^3 + (5/2)*(0)^2 + C*(0) + D
= 0 + 0 + 0 + 0 + D
D = 0
Therefore, the position function s(t) is given by:
s(t) = (1/12)t^4 - (4/3)t^3 + (5/2)t^2 + Ct
Step 3: Find the value of C using the condition s(1) = 20
Using s(1) = 20, we can substitute t = 1 and s(t) = 20 to find C.
20 = (1/12)*(1)^4 - (4/3)*(1)^3 + (5/2)*(1)^2 + C*(1)
= (1/12) - (4/3) + (5/2) + C
= 1/12 - 16/12 + 30/12 + C
= 15/12 + C
= 5/4 + C
C = 20 - 5/4
C = 80/4 - 5/4
C = 75/4
Therefore, the position function s(t) is given by:
s(t) = (1/12)t^4 - (4/3)t^3 + (5/2)t^2 + (75/4)t
Thus, we have found the position of the particle as a function of time.
To find the position of the particle, we need to integrate the acceleration function, a(t), twice with respect to time.
First, let's integrate a(t) to get the velocity function, v(t):
v(t) = ∫(a(t))dt = ∫(t^2 − 8t + 5)dt
To integrate each term, we use the power rule of integration:
∫(t^2)dt = (t^3)/3
∫(-8t)dt = -4t^2
∫(5)dt = 5t
Putting it all together, we have:
v(t) = (t^3)/3 - 4t^2 + 5t + C
Next, let's find the constant of integration, C, by using the initial condition s(0) = 0.
When t = 0, v(t) = 0, so we can plug these values into the velocity function:
0 = (0^3)/3 - 4(0)^2 + 5(0) + C
Simplifying, we find that C = 0.
Now, we have the velocity function:
v(t) = (t^3)/3 - 4t^2 + 5t
Finally, let's integrate v(t) to get the position function, s(t):
s(t) = ∫(v(t))dt = ∫((t^3)/3 - 4t^2 + 5t)dt
Again, using the power rule of integration, we have:
∫((t^3)/3)dt = (t^4)/12
∫(-4t^2)dt = (-4/3)t^3
∫(5t)dt = (5/2)t^2
Adding the integrals together and including the constant of integration, we get:
s(t) = (t^4)/12 - (4/3)t^3 + (5/2)t + C
Using the second initial condition, s(1) = 20, we can solve for the constant of integration, C:
20 = (1^4)/12 - (4/3)(1)^3 + (5/2)(1) + C
Simplifying, we find that C = 25/12.
Finally, substituting the value of C into the position function, we have the position of the particle:
s(t) = (t^4)/12 - (4/3)t^3 + (5/2)t + 25/12