Question

he position of a particle moving along an x axis is given by x = 13.0t² - 3.00t³, 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 = 7.00 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)? (i) Determine the average velocity of the particle between t = 0 and t = 7.00 s.

Answer 1

Position:

Velocity:

Acceleration:

a) t=7 seconds

Thus,

position x:

b) velocity v:

c) Acceleration a:

d)

Maximum positive coordinate reached by the particle:

condition: dx/dt = 0 or, v = 0 and d²x/dt² < 0

Thus Putting v = 0 we get:

or,

The roots of quadratic equation are:

or,

t = 0 or t = 26/9 s

At t = 0, a = 26 - 18t = 26 or, t >0 (condition not satisfied)

At t = 26/9 = 2.89 s, a = 26 - 52 = -26 m/s² or, a <0 (condition satisfied for maxima of x)

Hence, maximum positive coordinate is reached at t = 26/9 s = 2.89s

d) Maximum positive coordinate reached:

e) time at max cordinate reached: t = 26/9 = 2.89 seconds

f) Maximum positive velocity reached by the particle:

Condition: dv/dt = 0 or, a = 0

or, 26 - 18t = 0

or, 18t = 26

t = 26/18 s = 1.44s

Maximum velocity v =

g) time at which max velocity reached : t = 26/18 = 1.44 seconds (see above)

h) Acceleration of the particle at the instant when it is not moving.

When it is not moving, its velocity v=0

or,

From answer c) above this happens at two instants: t=0 and t=26/9 s

Thus acceleration at t=26/9 s :

At t = 26/9 = 2.89 s, a = 26 - 18*t= 26-18 * 2.89= 26 - 52 = -26 m/s²

i)

Average velocity of the particle between t =0 and t= 7s

From a) final distance (at t=7s) = -392 m

Assuming that the particle started from origin, x = 0 m

The average velocity = final diaplacement - initial diaplacement / (final time - initial time) = -392/7 = -56 m/s