Posted by **help** on Sunday, December 14, 2008 at 8:21pm.

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 -
**Damon**, Sunday, December 14, 2008 at 9:33pm
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

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