Chapter 4, Problem 55RE

(a)

To determine

The farthest apart the particles ever get.

Answer to Problem 55RE

The particles can be farthest apart at $0.7654$

Explanation of Solution

Given:

The position of two particles on the x-axis are $s_1=\cos t$ and $s_2=\cos (t+\frac{\pi }{4})$ .

Calculation:

For particles to be at farthest apart, s must be maximum therefore:

  $\begin{array}{l}d(t)=s_1(t)-s_2(t)\\ d(t)=cos(t)-cos(t+\frac{\pi }{4})\\ d\text{'}(t)=-\sin (t)+\sin (t+\frac{\pi }{4})\\ d\text{'}(t)=\sin (t+\frac{\pi }{4})-\sin (t)\\ d\text{'}(t)=2.\cos \left(\frac{t+t+\frac{\pi }{4}}{2}\right).\sin \left(\frac{t-(t+\frac{\pi }{4})}{2}\right)\\ d\text{'}(t)=2.\cos \left(t+\frac{\pi }{8}\right).\sin \left(-\frac{\pi }{8}\right)\\ d\text{'}(t)=0\\ \cos \left(t+\frac{\pi }{8}\right)=0\\ t+\frac{\pi }{8}=k\pi +\frac{\pi }{2}\\ t=k\pi +\frac{\pi }{2}-\frac{\pi }{8}\\ t=k\pi +\frac{3\pi }{8}\end{array}$

Now, check these at some critical points to find the farthest distance between particles.

  $\begin{array}{l}d\left(-1.\pi +\frac{3\pi }{8}\right)=-\sqrt{2-\sqrt{2}}\\ d\left(0.\pi +\frac{3\pi }{8}\right)=\sqrt{2-\sqrt{2}}\\ d\left(1.\pi +\frac{3\pi }{8}\right)=-\sqrt{2-\sqrt{2}}\\ d\left(2.\pi +\frac{3\pi }{8}\right)=\sqrt{2-\sqrt{2}}\end{array}$

As, the value of $\sqrt{2-\sqrt{2}}=0.7654$ Hence, the farthest distance between them is $0.7654$

Conclusion:

The particles can be farthest apart at $0.7654$

(b)

To determine

The distance at which the particle collides.

Answer to Problem 55RE

The distance at which the particle collides is $k\pi -\frac{\pi }{8}$ .

Explanation of Solution

Given:

The position of two particles on the x-axis are $s_1=\cos t$ and $s_2=\cos (t+\frac{\pi }{4})$ .

Calculation:

The distance at which particle collides can be find out as;

  $\begin{array}{l}{s}_1(t)=s_2(t)\\ \cos (t)=\cos (t+\frac{\pi }{4})\\ \cos (t)-\cos (t+\frac{\pi }{4})=0\\ -2.\sin \left(\frac{t+t+\frac{\pi }{4}}{2}\right).\sin \left(\frac{t-(t+\frac{\pi }{4})}{2}\right)=0\\ \sin \left(t+\frac{\pi }{8}\right).\sin \left(-\frac{\pi }{8}\right)=0\\ \sin \left(t+\frac{\pi }{8}\right)=0\\ t+\frac{\pi }{8}=k\pi \\ t=k\pi -\frac{\pi }{8}\end{array}$

Conclusion:

The distance at which the particle collides is $k\pi -\frac{\pi }{8}$ .