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.
4 answers
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.
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
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.
0 answers