The acceleration of a particle at a time t moving along the x-axis is give by: a(t) = 4e^(2t). At the instant when t=0, the particle is at the point x=2, moving with velocity v(t)=-2. Find the position of the particle at t=1/2

Question

Create a visual depiction of the concept related to physics and calculus in the given problem context. Depict a particle moving along an axis, showing acceleration and velocity indicators. Consider showing a curve that represents the particle's changing position over time, influenced by the accelerating function a(t) = 4e^(2t). Be sure to denote the starting point at t=0 with the x=2 position and v(t)=-2 velocity. Exclude the actual text of the problem or any mathematical solutions from the image.

The acceleration of a particle at a time t moving along the x-axis is give by: a(t) = 4e^(2t). At the instant when t=0, the particle is at the point x=2, moving with velocity v(t)=-2. Find the position of the particle at t=1/2

if you could show me how to get that please

Answer 1

a(t) = 4e^(2t)

so v(t) = 2e^(2t) + C (I integrated, since dv/dt = a)
when t=0, v=-2
-2 = 2e^0 + c
c = -4 and then
v)t) = 2e^(2t) - 4
since ds/dt = v
s = e^(2t) - 4t + k
when t=0, s=2 (I assume that is what you meant by x=2)
2 = e^0 - 4(0) + k
k = 1
then s(t) = e^(2t) - 4t + 1
s(1/2) = e^1 - 2 + 1
= e - 1

Answer 2

Well, let's start by finding the velocity function v(t) using the acceleration function a(t) given:

We know that acceleration is the derivative of velocity, so we can integrate a(t) to find the velocity function:

v(t) = ∫ a(t) dt = ∫ 4e^(2t) dt

To integrate e^(2t), we can use the substitution method. Let's set u = 2t, then du = 2dt, and we can rewrite the integral as:

v(t) = ∫ 4e^u * (1/2) du
v(t) = 2 ∫ e^u du
v(t) = 2 * e^u + C

Now, we know that at t = 0, the velocity v(t) is -2. So, let's substitute t = 0 into the equation:

-2 = 2 * e^(2 * 0) + C
-2 = 2 * e^0 + C
-2 = 2 * 1 + C
-2 = 2 + C
C = -4

Now, let's write v(t) in terms of t:

v(t) = 2 * e^(2t) - 4

To find the position function x(t), we know that velocity is the derivative of position. So, we can integrate v(t) to get x(t):

x(t) = ∫ v(t) dt = ∫ (2 * e^(2t) - 4) dt

Integrating the first term is similar to the previous step, and the integral of -4 (a constant) is just -4t.

x(t) = ∫ (2 * e^(2t) - 4) dt
x(t) = 2 * ∫ e^(2t) dt - 4 ∫ dt
x(t) = 2 * (1/2) * e^(2t) - 4t + D

Again, we need to determine D. At t = 0, the position x(t) is 2. Substituting these values:

2 = 2 * (1/2) * e^(2 * 0) - 4 * 0 + D
2 = 1 * 1 + D
2 = 1 + D
D = 1

Finally, we have the position function x(t):

x(t) = e^(2t) - 4t + 1

Now, we can find the position of the particle at t = 1/2 by substituting t = 1/2 into the equation:

x(1/2) = e^(2 * 1/2) - 4 * (1/2) + 1

Calculating this value, we get:

x(1/2) = e^1 - 2 + 1
x(1/2) ≈ 4.718

So, the position of the particle at t = 1/2 is approximately 4.718.

Answer 3

To find the position of the particle at t=1/2, we need to integrate the velocity function to get the position function.

Given:
a(t) = 4e^(2t) (acceleration function)
v(t) = -2 (velocity at t=0)
x(0) = 2 (position at t=0)

Step 1: Integrate the acceleration function to find the velocity function.
To integrate a(t), we can treat e^(2t) as a constant, as the derivative of e^(2t) is itself.

∫ a(t) dt = ∫ 4e^(2t) dt
= 4 * ∫ e^(2t) dt
= 4 * 1/2 * e^(2t) + C
= 2e^(2t) + C

Step 2: Find the constant of integration, C.
To find C, we use the given velocity at t=0.

v(t=0) = 2e^(2*0) + C
= 2 + C
= -2 (given)

Solving for C:
2 + C = -2
C = -4

So, the velocity function becomes:

v(t) = 2e^(2t) - 4

Step 3: Integrate the velocity function to find the position function.
To integrate v(t), we can treat e^(2t) as a constant, as the derivative of e^(2t) is itself.

∫ v(t) dt = ∫ (2e^(2t) - 4) dt
= 2 * ∫ e^(2t) dt - 4 * ∫ dt
= 2 * 1/2 * e^(2t) - 4t + C
= e^(2t) - 4t + C

Step 4: Find the constant of integration, C.
To find C, we use the given position at t=0.

x(t=0) = e^(2*0) - 4*0 + C
= 1 + C
= 2 (given)

Solving for C:
1 + C = 2
C = 1

So, the position function becomes:

x(t) = e^(2t) - 4t + 1

Step 5: Calculate the position at t=1/2.
To find the position at t=1/2, substitute t=1/2 in the position function:

x(t=1/2) = e^(2*(1/2)) - 4*(1/2) + 1
= e^1 - 2 + 1
= e - 1

Therefore, the position of the particle at t=1/2 is e - 1.

Answer 4

To find the position of the particle at t=1/2, we need to integrate the acceleration function to find the velocity function, and then integrate the velocity function to find the position function.

Given:
Acceleration function: a(t) = 4e^(2t)
Velocity function: v(t) = -2

Step 1: Integrate the acceleration function to find the velocity function.
∫ a(t) dt = ∫ 4e^(2t) dt

To integrate e^(2t), we use the rule:

∫ e^(kx) dx = (1/k) * e^(kx) + C

In this case, k = 2, so we have:

∫ 4e^(2t) dt = (4/2) * e^(2t) + C1
= 2e^(2t) + C1

Since we know the velocity at t=0 is -2, we can find the constant C1:

v(t) = 2e^(2t) + C1
v(0) = 2e^(2*0) + C1
-2 = 2 + C1
C1 = -4

Therefore, the velocity function is:

v(t) = 2e^(2t) - 4

Step 2: Integrate the velocity function to find the position function.
∫ v(t) dt = ∫ (2e^(2t) - 4) dt

To integrate 2e^(2t), we use the rule again:

∫ e^(kx) dx = (1/k) * e^(kx) + C

In this case, k = 2, so:

∫ 2e^(2t) dt = (2/2) * e^(2t) + C2
= e^(2t) + C2

And for -4, it is simply -4t + C3.

Therefore, the position function is:

x(t) = ∫ v(t) dt
= ∫ (2e^(2t) - 4) dt
= e^(2t) - 4t + C

Now we need to find the constant C using the initial condition at t=0, where x=2:

x(t) = e^(2t) - 4t + C
x(0) = e^(2*0) - 4*0 + C
2 = 1 + C
C = 1

Therefore, the position function is:

x(t) = e^(2t) - 4t + 1

Now we can find the position of the particle at t=1/2 by plugging in the value:

x(t=1/2) = e^(2*(1/2)) - 4*(1/2) + 1
= e^1 - 2 + 1
= e - 1

So, the position of the particle at t=1/2 is e - 1.

Answer 5

postition= integral v dt

v= int a dt
find v first, you know the constant of integration from at t=2

Then integrate again to get position.