Question

I have three questions I need assistance on please:

1. To make an object start moving on a surface with friction requires

(a) less force than it takes to keep it moving on the surface at constant velocity

(b) the same force as to keep it moving on the surface at constant velocity

(c) more force than it takes to keep it moving on the surface at constant velocity

(d) a force equal to the weight of the object

(2) At any time t greater than or equal to 0 s, an object moving along the x-axis has a postion given by

x(t)=9.0t-t^3 m.

(a) what are the object's velocity, v(t), and acceleration, a(t), at any time t?

(b) what are the object's position, velocity, and acceleration at t= 2.0 s?

(c) Between t=0 and t=4 s, what are the object's displacement, average velocity, and average acceleration?

(d) At what time(s),t, is the object at rest?

(3) The postion of an object (in m) at any time t is given by: r(t)= ( 1/3 t^3+t-2) i hap- 4t jhap

(a) what is the velcity of the object v(t)?

(b) what is the acceleration of the object a(t)?

(c) what are the instantaneous velocity and acceleration of the object at t=2s?

(d) what is the displacement of the object between t=0s and t=2s?

(e) what is the average velocity of the object between t=0s and t=2s?

(f) what is the average acceleration of the object between t=0s and t=2s?

(g) what is the average force on the object between t=0s and t=2s?

(h) what is the instantaneous force on the object at t=2s?

Answer 1

1) To make an object start moving on a surface with friction requires more force than it takes to keep it moving on the surface at constant velocity because  static friction is greater than kinetic friction

2) postion given by x(t)=9.0t-t^3 m.

   so(a). at any time t

v(t)=d(x(t))/dt = 9-3t^2 m/s

   a(t)=d(v(t))/dt =-6t m/s^2

(b). at t=2s

v(2)=9-(3 * 2^2)=-3 m/s

a(2)=-6 *2=-12m/s^2

(c). Between t=0 to t=4

the object's displacement=|x(4)-x(0)|

=|-28-0|=| -28|m=28 m

avg. velocity = $\frac{\int_{0}^{4}v(t)dt}{\int_{0}^{4}dt}$

= $\frac{\int_{0}^{4}(9-3t^2)dt}{\int_{0}^{4}dt}$

=-28/4

=-7 m/s

avg acceleration= $\frac{\int_{0}^{4}a(t)dt}{\int_{0}^{4}dt}$

= $\frac{\int_{0}^{4}(-6t)dt}{\int_{0}^{4}dt}$

=-48/4=12m/s^2

(3) The postion of an object (in m) at any time t is given by: r(t)= ( 1/3 t^3+t-2) i - 4t j

a).v(t)= [t^2 + 1]i -[4]j

b). a[t]=[2t]i

c). at t=2

v[2]=5i-4j

a[2]=4i

d). displacement between t=0 to t=2 assume at t=0 particle was at (0,0)

displacement in x direction =5 * 2= 10 m

in y direction =4*2=8m

so displacement is = (10^2 + 8^2 )^(.5)

=12.80 m

avg velocity = $\frac{\int_{0}^{2}v(t)dt}{\int_{0}^{2}dt}$

=[(14/3)i-8j]/2=(7/3)i - 4j

avg. acceleration = $\frac{\int_{0}^{2}a(t)dt}{\int_{0}^{2}dt}$ =2i

avg force will be equal to mass*av. acceleration

force will be equal to mass * acceleration =

Answer 2

1. Answer: Option - (c)

Reason: Static Friction is greater than kinetic friction

2. Answer: Give more points

3. Answer: Same as Above