A particle is moving with the given data. Find the position of the particle.
a(t) = 2t + 5, s(0) = 2, v(0) = −4
s(t) = ?
a(t) = 2t + 5
v(t) = t^2 + 5t + c
given: when t=0 velocity = -4
-4 = 0 + 0 + c
c = -4, so
v(t) = t^2 + 5t - 4
s(t) = (1/3)t^3 + (5/2)t^2 - 4t + k
when t = 0, s(0) = 2
2 = 0 + 0 + 0 + k, so k = 2, and finally
s(t) = ......
To find the position of the particle given the data, we can start by finding the velocity function and then integrating it to find the position function.
Given that a(t) = 2t + 5, and applying the initial condition v(0) = -4, we can find the velocity function v(t) using integration:
v(t) = ∫ (a(t)) dt = ∫ (2t + 5) dt = t^2 + 5t + C
Applying the initial condition v(0) = -4, we get:
-4 = (0)^2 + 5(0) + C
-4 = C
Therefore, the velocity function v(t) is:
v(t) = t^2 + 5t - 4
Next, we can find the position function s(t) by integrating the velocity function:
s(t) = ∫ (v(t)) dt = ∫ (t^2 + 5t - 4) dt
To integrate the function, we need to find the antiderivative of each term separately:
∫ t^2 dt = (1/3) t^3 + K1
∫ 5t dt = (5/2) t^2 + K2
∫ -4 dt = -4t + K3
Here, K1, K2, and K3 are constants of integration.
Combining the antiderivatives, we get:
s(t) = (1/3) t^3 + K1 + (5/2) t^2 + K2 - 4t + K3
Now, applying the initial condition s(0) = 2, we get:
2 = (1/3) (0)^3 + K1 + (5/2) (0)^2 + K2 - 4(0) + K3
2 = K1 + K2 + K3
Therefore, the position function s(t) is:
s(t) = (1/3) t^3 + (5/2) t^2 - 4t + 2
The position of the particle can be determined by plugging in the appropriate value of t into the position function.
To find the position of the particle, we can integrate the acceleration function to get the velocity function, and then integrate the velocity function to get the position function.
Step 1: Integrate the acceleration function to get the velocity function.
∫a(t) dt = v(t)
∫(2t + 5) dt = v(t)
t^2 + 5t + C1 = v(t)
Step 2: Use the initial velocity condition to find the constant of integration.
v(0) = -4
0^2 + 5(0) + C1 = -4
C1 = -4
Therefore, the velocity function is:
v(t) = t^2 + 5t - 4
Step 3: Integrate the velocity function to get the position function.
∫v(t) dt = s(t)
∫(t^2 + 5t - 4) dt = s(t)
1/3 * t^3 + 5/2 * t^2 - 4t + C2 = s(t)
Step 4: Use the initial position condition to find the constant of integration.
s(0) = 2
1/3 * 0^3 + 5/2 * 0^2 - 4 * 0 + C2 = 2
C2 = 2
Therefore, the position function is:
s(t) = 1/3 * t^3 + 5/2 * t^2 - 4t + 2
So, the position of the particle at any time t can be found using the position function s(t) = 1/3 * t^3 + 5/2 * t^2 - 4t + 2.