Basic physics question!! Need Help!!!!?

Question

The position of a particle moving along an x axis is given by x = 15.0t^2 - 4.00t^3, where x is in meters and t is in seconds. Determine (a) What is the maximum positive coordinate reached by the particle and (b) at what time is it reached? (c) What is the maximum positive velocity reached by the particle and (d) at what time is it reached? (e) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (f) Determine the average velocity of the particle between t = 0 and t = 3.00 s

Answer 1

I assume you are using calculus in your course. If so:

a. Max coordinate occurs when dx/dt = 0 = 30t -12 t^2

Solving for t ---> 0 = 5 - 2t ---> t = 5/2

Then x = 15*(5/2)^2 - 4*(5/2)^3 you can plug this nto your calculator.

b. t = 5/2 - see above

c. v = dx/dt = 30 t - 12 t^2

Max velocity occurs when dv/dt = 0 = 30 - 24 t ---> t = 30/24 = 5/4

So vmax = 30*(5/4)-12*(5/4)^2 again you can do the arithmetic

d. t= 5/4 (see above)

e. The particle is not moving when dx/dt =0 = 30t - 12t^2 and we know the value of t from part b.

Acceleration a = dv/dt = 30 - 24t and t = 5/2

so a = 30 - 24*(5/2) = 30 -60 =-30

f. Average velocity = displacement divided by time

x(t=0) = 0, x(t=3) = 15*9 - 4*27 --> = (15*9 - 4*27)/3

= 45 - 36 = 11 unit lengths/s (don't know what units you are using)

Answer 2

x' = velocity = 30t - 12t^2
x'' = acceleration = 30 - 24t

a and b) maximum positive coordinate indicates a maxium of the displacement function. also, the velocity should be 0 since when it turns negative, the particle will be traveling back to the start

x' = 0 = 30t - 12t^2
0 = 3t(10-4t)
t = 10/4 = 2.5 s

plug it back into your function
x(2.5) = 15*2.5^2 - 4*2.5^3 = 31.25 (you didn't give us your units for distance)

c and d) the maxium positive velocity occurs before the acceleration goes negative

a = 0 = 30-24t
t = 1.25 s

v = 30*1.25 - 12*1.25^2 = 18.75

e) x'' = acceleration = 30 - 24t
the velocity = 0 at t=2.5
a = 30-24*2.5 = -30

f) avg velocity = final displacement/total time = 27/3 = 9