The graph give the acceleration a versus time t for a particle moving along an x axis. The a axis scale set by a=12 m/s^2. At t = -2s the particle's velocity is 10 m/s. What is its velocity at t = 6s.

I know that I am supposed to used integrals to solve, but I can't figure it out. a(t) = -2x+8

If someone could show me the steps then I could find the answer.

To find the velocity of the particle at t = 6s, you need to integrate the given acceleration function with respect to time.

Given: a(t) = -2t + 8

Step 1: Integrate the acceleration function to find the velocity function.
The integral of a(t) with respect to t gives the velocity function v(t). The constant of integration will be determined using the initial velocity at t = -2s.

∫a(t) dt = ∫(-2t + 8) dt

Using the power rule of integration, we can integrate each term separately:
∫(-2t) dt = -t^2
∫8 dt = 8t

Combining the results, we have:
v(t) = -t^2 + 8t + C

Step 2: Determine the constant of integration.
We can determine the constant of integration (C) by using the initial condition when t = -2s and v = 10m/s.

v(-2) = -(-2)^2 + 8(-2) + C = 10

Simplifying the equation:
4 - 16 + 8 + C = 10
-4 + C = 10
C = 10 + 4
C = 14

Step 3: Substitute the constant of integration into the velocity function.
Now we have the velocity function:
v(t) = -t^2 + 8t + 14

Step 4: Find the velocity at t = 6s.
Substitute t = 6 into the velocity function:
v(6) = -(6)^2 + 8(6) + 14
= -36 + 48 + 14
= 26

Therefore, the velocity of the particle at t = 6s is 26 m/s.

To find the velocity at t = 6s, you need to integrate the acceleration function over the time interval from -2s to 6s.

First, let's integrate the given acceleration function, a(t) = -2t + 8, with respect to time:

∫(a(t)) dt = ∫(-2t + 8) dt

To integrate -2t, we need to use the power rule of integration, which states that ∫x^n dx = (1/(n+1))x^(n+1) + C, where C is the constant of integration.

So, integrating -2t with respect to t, we get: -t^2

To integrate the constant term 8, we can simply multiply it by t, as the integral of a constant is equal to the constant multiplied by the variable of integration.

Therefore, the integral of 8 with respect to t is 8t.

Now, let's integrate the acceleration function over the time interval [-2s, 6s]:

∫[a(t)] dt = ∫[(-2t + 8) dt]

Integrating the function, we have: ∫[(-2t + 8) dt] = (-t^2 + 8t) evaluated from -2 to 6

Substituting the upper and lower limits of integration, we have: [(-6^2 + 8*6) - (-(-2^2) + 8*(-2))]

Simplifying further: [(-36 + 48) - (4 - 16)]

Finally, evaluating the expression: [12 - (-12)] = 24 m/s

Therefore, the particle's velocity at t = 6s is 24 m/s.

I don't know what you mean when you say: "The a axis scale set by a = 12 m/s^2."

That is not a sentence, and there is no "a axis".

t = 6s is 8 seconds after t = -2 s. If the acceleration rate is constant at 12 m/s^2, the velocity will increase by 8*12 = 96 m/s during that interval. That would make the new velocity 106 m/s.