s = 3t^2 – t^3 = 0

Factor it:

t^2(3 – t) = 0

So you have a double-root at t = 0 and a root at t = 3. It will be at s = 0 twice: once when the problem starts (t = 0) and again when t = 3.

2) The function is cubic in t – it drops to 0 for all t <=0, goes up and comes back down from t = 0 to 3, and continues down for the rest of history. During the fifth second, then, it moves in only one direction. To get the distance travelled during the fifth second, all you need to do is take s(6) - s(5), the postion at t = 6 and subtract the position at t = 5. s(6) = 3(6^2) - 6^3 = 108 - 216 = -108 So it travels 108 meters in the negative direction. 3) We know that it turns around once during the first 4 seconds (because it comes back to 0 at t = 3). To get the total distance, we need to know where and when it turned around. Take the first derivative, the velocity, and set it to 0. That'll be when it turned around: s'(t) = 6t - 3t^2 = 0 3t(2 - t) = 0 t = 0, which we already knew from the shape, and t = 2 So it travels in the positive direction from 0 to 2 and in the negative direction from 2 forever after. To get the total distance for the first four seconds, we take the distance for the first two seconds (s(2) - s(0)) and add the absolute value of the distance traveled during the next two seconds, s(4) - s(2). s(0) = 0 s(2) = 3(2^2) - 2^3 = 12 - 8 = 4 s(4) = 3(4^2) - 4^3 = -16 So the total distance during the first two seconds is 4m. The total distance during the second two seconds is 20m (from +4 to -16). All together, it travels 24m during the first four seconds. 4) The maximum velocity will occur when the velocity has a critical point: in other words, where the second derivative (the acceleration) is 0 or undefined. The second derivative is a(t) = v'(t) = s''(t) = d/dt(6t - 3t&2) (using the velocity function we got above) a(t) = 6 - 6t That will equal 0 at t = 1, at which point the velocity will be v(1) = 6(1) - 3(1)^2 = 3 m/s Note, please, that this is the maximum velocity, NOT the maximum speed. It's the maximum velocity because it's the fastest it'll be going in the positive direction. In the negative direction, there will be no maximum. You should be able to see why. Think of the slope of the position graph (the s(t)), which is the velocity. I know this is long. Hope it helped.