Question

The position of a particle moving along an x axis is given by x = 14.0t2 - 2.00t3, 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 = 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)? (i) determine the average velocity of the particle between t = 0 and t = 5.00 s.

Answer 1

a) x = $14t^{2} -2t^{3}$

at t = 5s

$x = 14*5^{2} -2*5^{3} = 14*25 - 2*125 = 350 - 250 = 100 m$

b) v = $\frac{dx}{dt}$

       = $28t - 6t^2$

at t = 5 s

v = $28*5-6*25 = 140 - 150 = -10 m/s$

c) a = $\frac{dv}{dt}$

     = 28 - 12t

at t = 5 s

a = 28 -12*5= 28-60= -32 m/$s^2$

d) At maximum positive coordinate velocity = 0

So, $0 = 28t - 6t^2$

         $t = \frac{28}{6} = \frac{14}{3} = 4.66 s$

  At t = 4.66 s

   $X = 14 * 4.66^2 - 2* 4.66^3 = 202.39 m$

e) At t = 4.66 s

f) At maximum positive velocity a = 0

   $0=28-12t$

   $t = \frac{28}{12} = \frac{7}{3} = 2.33 s$

At t = 2.33 s

V = $28*2.33- 6*2.33^2= 32.67 m/s$

g) t = 2.33 s

h) When particle is not moving v = 0

So $0= 28t - 6 t^2$

 $t = \frac{28}{6} = 4.66 seconds$

At t = 4.66 s

a = $28 - 12 * 4. 66 = -27.93m/s^2$

i) At t = 0s, X =0m

       t = 5s, X = 100m

So, Displacement = 100m

Velocity = $\frac{Displacement}{Time} = \frac{100}{5} = 20m/s$