# physics

posted by on .

The position of a particle moving along an x axis is given by x = 15t2 - 2.0t3, where x is in meters and t is in seconds.
(a) Determine the position, velocity, and acceleration of the particle at t = 3.0 s.
x = m
v = m/s
a = m/s2

(b) What is the maximum positive coordinate reached by the particle?
m
At what time is it reached?
s

(c) What is the maximum positive velocity reached by the particle?
m/s
At what time is it reached?
s

(d) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)?
m/s2

(e) Determine the average velocity of the particle between t = 0 and t = 3 s.
m/s

• physics - ,

x = 15t^2 - 2.0t^3
v = dx/dt = 30 t - 6 t^2
a = d^2x/dt^2 = 30 - 12 t

part a, just put in 3 for t in the above

part b
max position when v = 0
0 = t (30 - 6t)
so at t = 5 sec
then
x = 15*25 - 2*125 = 125

part c
max v when a = 0
t = 30/12
v = 30 (30/12) - 6 (30/12)^2

part d
not moving at t = 5 from part b
a = 30 - 12(5)

part e
Vmean (3) = x at 3 - x at 0
Vmean = [15(9) - 2.0(27) - 0 ]/3