a particle moves along a straight line and has an acceleration given by a(t) = 6t+4 where a is measured in cm/sec^2. Its initial velocity is -6 cm/sec. Initially it sits 9 cm to the right of the origin. Find its displacement function?
Can anyone show me what i need to do for this, thanks
assuming to the right is positive, then we have
a = 6t+4
v = 3t^2+4t+c
v(0) = c = -6
s = t^3+2t^2-6t+c
s(0)=c=9
s(t) = t^3+2t^2-6t+9
To find the displacement function of the particle, follow these steps:
Step 1: Integrate the acceleration function to find the velocity function.
The acceleration function is given as a(t) = 6t + 4. Integrating this function will provide the velocity function.
Integrating a(t) will give you:
v(t) = ∫ (6t + 4) dt
The integral of 6t becomes (6/2) * t^2 = 3t^2, and the integral of 4 becomes 4t.
So, v(t) = 3t^2 + 4t + C, where C is the constant of integration.
Step 2: Use the initial velocity to determine the constant of integration.
The particle's initial velocity is given as -6 cm/sec. Plug this value into the velocity function obtained in step 1, and solve for C.
v(0) = 3(0)^2 + 4(0) + C
-6 = C
C = -6
Therefore, the velocity function becomes v(t) = 3t^2 + 4t - 6.
Step 3: Integrate the velocity function to find the displacement function.
Integrating the velocity function will provide us with the displacement function.
∫ (3t^2 + 4t - 6) dt = t^3 + 2t^2 - 6t + D, where D is the constant of integration.
Step 4: Use the initial displacement to determine the constant of integration.
Initially, the particle sits 9 cm to the right of the origin (at t = 0). Substituting t = 0 into the displacement function and equating it to the initial displacement will allow us to find the value of D.
D(0) = (0)^3 + 2(0)^2 - 6(0) + D
0 = D + 0
0 = D
Therefore, the constant of integration D is 0.
Finally, the displacement function of the particle is:
s(t) = t^3 + 2t^2 - 6t
This equation represents the position of the particle as a function of time on the straight line.
To find the displacement function, we need to integrate the velocity function over time. In this case, we first need to find the velocity function.
Given that the acceleration function is a(t) = 6t + 4 and the initial velocity is -6 cm/sec, we can integrate the acceleration to find the velocity function.
To integrate the acceleration function a(t), we integrate each term separately. Keep in mind that the integral of a constant is the constant multiplied by the variable.
∫(6t + 4) dt = ∫6t dt + ∫4 dt
Integrating each term separately:
= 3t^2 + 4t + C
Here, C is the constant of integration.
We are given that the initial velocity is -6 cm/sec, so we can substitute the initial time t = 0 into the velocity function and solve for the constant of integration.
-6 = 3(0^2) + 4(0) + C
-6 = 0 + 0 + C
C = -6
Substituting C back into the velocity function, we have:
v(t) = 3t^2 + 4t - 6
Now, we can integrate the velocity function to find the displacement function. The displacement function is the integral of the velocity function with respect to time.
∫(3t^2 + 4t - 6) dt = ∫3t^2 dt + ∫4t dt - ∫6 dt
Integrating each term separately:
= t^3 + 2t^2 - 6t + K
Here, K is the constant of integration.
As we are given that the particle initially sits 9 cm to the right of the origin, we can substitute the initial time t = 0 into the displacement function and solve for the constant of integration.
9 = (0^3) + 2(0^2) - 6(0) + K
9 = 0 + 0 + 0 + K
K = 9
Substituting K back into the displacement function, we have:
d(t) = t^3 + 2t^2 - 6t + 9
Therefore, the displacement function of the particle in this scenario is d(t) = t^3 + 2t^2 - 6t + 9.