Question

Question #5c9ea

Answer 1

`-50color(white)(l)"km/h"color(white)(l)hati - 50color(white)(l)"km/h"color(white)(l)hatj`

Explanation:

First, let's find the components of the initial velocity when it was traveling upward:

• `v_(0x) = 0` (motion is solely along `y`-axis)

• `v_(0y) = 50` `"km/h"`

A `90^"o"` left turn means that it would now be traveling in the negative `x`-direction (i.e. west ) with the same speed of `50` `"km.h"`. The final velocity components are thus

• `v_(1x) = -50` `"km/h"`

• `v_(1y) = 0` (motion is now solely horizontal)

The change in velocity `Deltavecv` is

`color(red)(Deltavecv) = overbrace(vecv_1)^"final velocity" - overbrace(vecv_0)^"initial velocity" = (-50color(white)(l)"km/h"color(white)(l)hati + 0hatj) - (0hati + 50color(white)(l)"km/h"color(white)(l)hatj)`

`= color(red)(ulbar(|stackrel(" ")(" "-50color(white)(l)"km/h"color(white)(l)hati - 50color(white)(l)"km/h"color(white)(l)hatj" ")|)`