A particle is moving with the given data. Find the position of the particle.
a(t) = t2 − 5t + 3, s(0) = 0, s(1) = 20
s(t) =
To find the position function, s(t), we need to integrate the given acceleration function, a(t).
Given: a(t) = t^2 - 5t + 3
To integrate a(t), we use the power rule for integration:
∫ (t^n) dt = (t^(n+1))/(n+1) + C
Integrating a(t):
∫ (t^2 - 5t + 3) dt = (∫ t^2 dt) - (∫ 5t dt) + (∫ 3 dt)
Using the power rule:
= (t^(2+1))/(2+1) - (5/2) * (t^(1+1))/(1+1) + (3t)
Simplifying:
= (t^3)/3 - (5/2) * (t^2)/2 + (3t)
Now, we need to find the position function using the initial conditions.
Given: s(0) = 0
Using this condition, when t = 0, s(t) should be equal to 0.
Plugging t = 0 into the above equation:
s(0) = (0^3)/3 - (5/2) * (0^2)/2 + (3*0)
0 = 0 + 0 + 0
0 = 0
Given: s(1) = 20
Using this condition, when t = 1, s(t) should be equal to 20.
Plugging t = 1 into the equation:
s(1) = (1^3)/3 - (5/2) * (1^2)/2 + (3*1)
= 1/3 - (5/2) * 1/2 + 3
= 1/3 - 5/4 + 12/4
= (4/12) - (15/12) + (48/12)
= 37/12
Therefore, the position function, s(t), is:
s(t) = (t^3)/3 - (5/2) * (t^2)/2 + (3t)
Alternatively, if we simplify the equation:
s(t) = (t^3)/3 - (5/4) * (t^2) + (3t)
To find the position of the particle, we need to integrate the given acceleration function, a(t), twice with respect to time (t). The first integral will give us the velocity function, and the second integral will give us the position function.
Step 1: Integrate the acceleration function, a(t), to get the velocity function, v(t).
To integrate a(t), we will treat it as a polynomial expression and apply the power rule of integration:
∫(t^2 - 5t + 3) dt
Applying the power rule, we get:
= (1/3)t^3 - (5/2)t^2 + 3t + C
Let's call this integrated function as V(t), which represents the velocity function.
Step 2: Find the constant of integration, C.
To determine the value of C, we can use the initial condition s(0) = 0, which means the particle starts at position 0 when t = 0.
We know that velocity is the rate of change of position, so v(t) = ds(t)/dt. We can express this relationship mathematically as:
v(t) = d/dt [s(t)]
Integrating both sides of the equation, we get:
∫[v(t)] dt = ∫[d/dt [s(t)]] dt
This simplifies to:
∫[v(t)] dt = ∫ ds(t)
V(t) = s(t) + C
Since s(0) = 0, when t = 0 in the velocity function V(t), we have:
V(0) = s(0) + C
0 = 0 + C
C = 0
Therefore, the constant of integration, C, is 0.
Step 3: Integrate the velocity function, V(t), to get the position function, s(t).
To integrate V(t), we will treat it as a polynomial expression and apply the power rule of integration:
s(t) = ∫[V(t)] dt
= ∫[(1/3)t^3 - (5/2)t^2 + 3t] dt
Applying the power rule, we get:
s(t) = (1/12)t^4 - (5/6)t^3 + (3/2)t^2 + C
Since the constant of integration, C, is 0 in this case, we can simplify the expression to:
s(t) = (1/12)t^4 - (5/6)t^3 + (3/2)t^2
This is the position function of the particle, s(t), in terms of time, t.
a = t^2-5t+3
v = 1/3 t^3 - 5/2 t^2 + 3t + C1
s = 1/12 t^4 - 5/6 t^3 + 3/2 t^2 + C1*t + C2
Now use the conditions to solve for C1 and C2.