A particle moves along the x-axis so that its velocity at time t is given by...?

Question

V(t) = -(t+1)(sin(t^2/2))
at t=0, the particle at position x=1.

A> find the acceleration of the particle at t=2. Is the speed of the particle increasing at that time why or why not?

B> find all the times t in the ioen interbal 0 < t < 3 when the particle changes direction, explain.

C> find the total distance traveled by the particle from t=0 to t=3.

D> during the time interbal between t greater than or equal to 0 but less than or equal to three, what is the greatest distance between the particle and the origin? explain.

Answer 1

V(t) = -(t+1)(sin(t^2/2))
a




V(t) = -(t+1)(sin(t^2/2))
a(t=(sin(t^2/2))-(cos(t^2/2))*t(t+1).
a(t=2)=sin(2)-6cos(2).=+, because, in the second quadrant, sin=+and cos=-so the speed isincreasing at t=2s.
v=0 when -(t+1)sin(t^2/2))=0,
when t^2/2=pi/2 or 3pi/2
or t=sqrt(2pior sqrt(3pi).
the particle changes direction when t=sqrt(2pi)=.4.44s.

V(t) = -(t+1)(sin(t^2/2))
x=-(t+1)^2/2 sin(t^2/+cos(t^2/2)

Answer 2

acceleration= dV/dt= -[sin(t^2/2)+( t-1) cos(t^2/2)*t ]At t=2
a= -(sin2+cos2*2]=- 0.0770 negative ,so the velocity is DECREASING.
B) You have to study the sign of V(t)
-(t+1) has a constan sign in 0
the sign of sin(t^2/2)
if t^2/2 0 At t=2.4912 the particle changges direction
The other time of change of direction would be t =sqrt(4pi) =3.5231 outside the interval.
C) d= Int (t+1)sin(t^2/2) dt (0,2.4912) -Int(t+1)sin(t^2/2)dt (2.4912,3)

Let´s calculate Int (t+1)sin(t^2/2) dt = Int t*sin(t^2/2)dt+ Int sin(t^2/2)dt

t^2/2=z so t dt =dz and the first int becomes Int sinz dz=
-cos(t^2/2) (between 0,and sqrt(2pi) = 2+toward the origin as velocity is negative+Intsin(t^2/2)dt (0,sqrt(2pi))
Frankly I could not calculate the integral of sin(t^2/2)dt

Answer 3

A> a = d/dt(v(t)) = - sin(t^2/2) - (t+1)cos(t^2/2)*t
t=2 = -sin(2) - 3cos(2)*2
= - sin(2*180/pi) - 3cos(2*180/pi
)*2
= - sin(114.6deg) - 3cos(114.6 deg)*2
= - sin ( 65.4deg) - 3(sin 25.4)*2
a is in the same dir. as v. hence v is increasing

p.s. do not trust me, it's quite late in the night here, good luck