I am given two points on an acceleration vs time graph: (-2,13),(4,0). When t = -2s, v = 10 m\s. What velocity at t = 6s? I believe acceleration average is (-13/6), but how do I solve the rest get the correct answer, which is 44.7 m/s? Any help or hints would be greatly appreciated.

If v=10 when t= -2, and you have negative acceleration, v is decreasing. How can it increase to 44.7? Is there a typo in here somewhere?

Never mind. I see acceleration is not constant.

However, I get 15.3 m/s. I will have to recheck my math.

OK. Finally got a moment,

If we assume a linear function for acceleration, and fitting it to the two points, we get

a = -13/6 t + 26/3

v = V0 + 26/3 t - 13/12 t2

v(2) = 10 = V0 - 26/3 * 2 - 13/12 * 4
10 = V0 -65/3
V0 = 95/3

So, now we know

v(t) = 95/3 + 26/3 t - 13/12 t2
v(6) = 95/3 + 26/3 * 6 - 13/12 * 36 = 44 2/3 = 44.7

An autos velocity increases uniformly from 6m/s to20m/s while covering 70m.find the acceleration and the time taken.

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

First, let's find the equation of the acceleration function given the two points (-2,13) and (4,0). We can use the formula for the equation of a line, y = mx + b, where m is the slope and b is the y-intercept.

The slope, m, can be calculated as the change in velocity divided by the change in time. In this case, the change in velocity is 0 - 13 = -13, and the change in time is 4 - (-2) = 6. So, the slope, m, is -13/6.

Now we have the equation for the acceleration function: a(t) = (-13/6)t + b.

Next, let's use the given information at t = -2s: v = 10 m/s.

The velocity, v, is the integral of the acceleration function over the time interval from -2s to t. So we need to find the antiderivative (or integral) of a(t) and solve for b.

Integrating a(t), we get v(t) = (-13/12)t^2 + b*t + C, where C is the constant of integration.

Plugging in t = -2s and v = 10 m/s, we get 10 = (-13/12) * (-2)^2 + b * (-2) + C.

Simplifying the equation, we have 10 = 26/3 - 2b + C.

Solving for C, we subtract 26/3 from both sides and multiply by -1, so C = -8/3 + 2b.

Now, we have the equation for the velocity function: v(t) = (-13/12)t^2 + b*t - (8/3 - 2b).

Lastly, to find the velocity at t = 6s, we plug in t = 6s into the velocity function and solve for v.

v(6) = (-13/12)*(6)^2 + b*(6) - (8/3 - 2b).

Simplifying the equation, we have v(6) = -78/12 + 6b - (8/3 - 2b).

Combining like terms, we get v(6) = -39/6 + 10b/3 + 8/3.

To find b, we can use the average acceleration formula you mentioned: acceleration average = (-13/6).

The average acceleration is the change in velocity divided by the change in time. In this case, the change in time is 6 - (-2) = 8. So, the change in velocity is 0 - 13 = -13. Therefore, (-13/6) = (v(6) - 10)/(8).

Solving for v(6), we have -13/6 * 8 = v(6) - 10, v(6) - 10 = -26/3.

Adding 10 to both sides, we get v(6) = -26/3 + 10, v(6) = 44/3.

Converting to decimal form, v(6) is approximately 14.67 m/s, which is not the correct answer.

It seems there might be a mistake or inconsistency in the given information or calculations. Do double-check the inputs and calculations used to verify the correctness of the answer.