Asked by Belinda

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.
Answered by Steve
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?
Answered by Steve
Never mind. I see acceleration is not constant.

However, I get 15.3 m/s. I will have to recheck my math.
Answered by Steve
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 = V<sub>0</sub> + 26/3 t - 13/12 t<sup>2</sup>

v(2) = 10 = V<sub>0</sub> - 26/3 * 2 - 13/12 * 4
10 = V<sub>0</sub> -65/3
V<sub>0</sub> = 95/3

So, now we know

v(t) = 95/3 + 26/3 t - 13/12 t<sup>2</sup>
v(6) = 95/3 + 26/3 * 6 - 13/12 * 36 = 44 2/3 = 44.7
Answered by Anonymous
An autos velocity increases uniformly from 6m/s to20m/s while covering 70m.find the acceleration and the time taken.
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