A 60 N block in horizontal plane is pushed horizontally by force P = t^3 + 10t. The frictional resistance f = 6 - t^2. What is the distance traveled by the block after 2 seconds.
The net force F = (t^3 + 10t) - (6 - t^2) = t^3 + t^2 + 10t - 6.
The Attempt at a Solution
There are actually three questions in this problem, acceleration, velocity and distance, all after 2 seconds. I encountered no problem with the acceleration which is from F = Ma
a = (g/W) (t^3 + t^2 + 10t - 6) and after 2 seconds, a = 2.398 m/s
Also no problem with the velocity by taking dv / dt = a, it is straightforward to integrate dv = a dt from known acceleration equation above. The equation of velocity is v = (g/W) [ (t^4 / 4) + (t^3 / 3) + 5t^2 - 6t ].
My question is the distance, with equation ds = v dt.
I have difficulty deciding which one to take for the initial time. Since at t = 0, the frictional resistance f is greater than the horizontal force P. I initially took t from zero to 2, but found that motion will start to impend not until 0.552 590 6474 seconds. Integration to distance equation will yield 0.7 for limits from 0 to 2 and 0.8 if limits is from 0.552 590 6474 seconds to 2 seconds. So which one is the right answer, I need your opinion. Thanks.