I don't like this question. The position of a particle moving along an x axis is given by x = 15.0t^2 - 3.00t^3, where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t = 5.00 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)?

Question

anonymous · Answer

just derivative and integral. try it

anonymous · Answer

Physics and any sort of math is not my specialty... could you say more?

anonymous · Answer

ok wait

anonymous · Answer

a) $$x=15.t^{2}-3.t^{3}$$  (t=5) 
$$x=15.5^{2}-3.5^{3}$$
x=300

b) derivative  (t=5)
$$x =15.t ^{2}-3.t ^{3}$$
$$V =30.t-9.t ^{2}$$
V= -74

c) derivative again  (t=5)
$$x =15.t ^{2}-3.t ^{3}$$
$$V =30.t-9.t ^{2}$$
$$a=30-18t$$
a=-60

part 1

anonymous · Answer

it says "What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0) "  that means the particle is slowing down and some time later it will stop :)) and for finding when it will stop, you should use $$V=30.t-9.t ^{2}$$  (V=0 when stops)