Question

Question 2. An object travels with velocity (2,3,5) m/s. (a) What is the speed of the object? (b) Find a unit vector in the direction of (2,3,5), and one in the opposite direction. (c) The object slows down to speed 3 m/s without changing direction Give its velocity vector.

Answer 1

Answer:

Step-by-step explanation:

Part 1)

Since speed defined as  the magnitude of velocity vector we have

$|v|=\sqrt{v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}$

Applying values we get

$|v|=\sqrt{2^{2}+3^{2}+5^{2}}$

$\therefore v=6.16m/s$

Thus the speed of the object is 6.16 m/s

Part 2)

A unit vector in the direction of $\overrightarrow{v}=v_{x}\widehat{i}+v_{y}\widehat{j}+v_{z}\widehat{k}$ is given by

$\frac{v_{x}\widehat{i}+v_{y}\widehat{j}+v_{z}\widehat{k}}{|v|}$

Applying values we get

Unit vector in direction of the given vector is

$\frac{2\widehat{i}+3\widehat{j}+5\widehat{k}}{6.16}\\\\0.324\widehat{i}+0.487\widehat{j}+0.811\widehat{k}$

The unit vector in opposite direction is

$-0.324\widehat{i}-0.487\widehat{j}-0.811\widehat{k}$

Part 3)

Since the speed of the particle reduces to 3 m/s from 6.16 m/s this can be brought about by varying each $v_{x},v_{y},v_{z}$ thus the problem is indeterminate since the no. of variables is 3 ( each individual component of velocity can be changed independently) but the equation avaliable is only 1. Thus we cannot find the velocity vector only by the information given.