2006-07-24 12:24:22 · 8 answers · asked by tikki 2 in Science & Mathematics ➔ Mathematics
I don't know about distance, but if t^3-2t refers to the position of the particle at time t, the acceleration would be the second derivative of that, which is 6t.
2006-07-24 12:27:54 · answer #1 · answered by Vic 2 · 2⤊ 2⤋
Hmm, the t^3 is odd because this means that acceleration cannot be constant but doing the maths....
s=t^3 -3t
ds/dt = 3t^2 -2
d2s/dt2 = 6t
A strange situation, acceleration is increasing with time. Hope that helps with your homework!
2006-07-24 19:41:03 · answer #2 · answered by mattpa 1 · 0⤊ 0⤋
Take the derivative twice
ds/dt = 3t^2-2
ds/dt^2 = 6t
2006-07-24 19:32:24 · answer #3 · answered by Michael M 6 · 0⤊ 0⤋
distance = s = t^3 - 2t
velocity = s' = 3t^2 - 2
acceleration = s'' = 6t
2006-07-25 00:17:31 · answer #4 · answered by Anonymous · 0⤊ 0⤋
s=t^3-2t
we know
v=ds/dt
v=3t^2-2
a=dv/dt;
a=6t; ans
2006-07-24 19:33:07 · answer #5 · answered by Gunjit M 2 · 0⤊ 0⤋
that's the second derivate...
s' = 3t^2 - 2
s''= 6t
2006-07-24 20:44:02 · answer #6 · answered by Anonymous · 0⤊ 0⤋
aceleration of -4m/s^2
or retardation of 4m/s^2
2006-07-25 11:50:23 · answer #7 · answered by a tactician 1 · 0⤊ 0⤋
Some other formula.
2006-07-24 19:27:05 · answer #8 · answered by dopeysaurus 5 · 0⤊ 0⤋