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.
a(t) = t2 − 5t + 3, s(0) = 0, s(1) = 20
s(t) =
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.
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)
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.