Question #5c9ea

1 Answer
Nathan L.
Aug 23, 2017

#-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" ")|)#

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