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

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}}`

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.