Question

(15) 6) A particle moves with acceIeration given by a(t) = COS(t) - sin (t) m /s^2 on the interval [0, 5]. Assume the initial velocity is I m /s and the initial position is at the origin. a) Find the velocity and position functions. b) When is the velocity zero? Positive? Negative? c) What is the maximum and minimum values of position on this interval?

Answer 1

a(t)=cos t-sin t

dv/dt=cos t-sin t

integrating,
v(t)=sin t+cos t +c

where c is a constant

now a t=0,
v=1

so c=0

v(t)=sin t+cos t

dx/dt=sin t+cos t

x(t)=-cos t+sin t+c

x(0)=0

so c-1=0

c=1

hence x(t)=sin t-cos t+1

b)v(t)=sin t+cos t

v(t)=0

==> sin t=-cos t

tan t=-1

t=-pi/4 or 3*pi/4

c)x(t)=sin t-cos t+1

plotting it for t=[0,5] we can see that

minimum occurs t at=5 and maximum occurs at dx/dt=0

i.e. t=3*pi/4=2.356 seconds

minimum value=-0.24259

maximum value=2.4142