Chris drives from Ottawa to Huntsville $(342$ km$),$ then from Huntsville to Toronto $(236$ km$).$ 

a$)$ What is the magnitude and direction of Chris$’$ displacement in km$?$ 

b$)$ If the trip takes $4\mathrm{.}75$ hours, what is the magnitude of Chris$’$ average velocity in km$/$h$?$

c$)$ c$)$ At one instance during the drive between Huntsville and Toronto, Chris$’$ velocity changes from $-30\mathrm{.}56$ m$/$s to $-41\mathrm{.}44$ m$/$s in $19\mathrm{.}60$ s $($the negative signs indicate that he is driving southwards, or down$).$ What is his average acceleration over the $19\mathrm{.}60$ s $($keep your units in m and s; do not convert to km and h$)?$ 

d$)$ How is the sign $($i$.$e$.,$ positive or negative$)$ of the acceleration calculated in part c$)$ interpreted? That is$,$ which of the $4$ interpretations from the table below $($from lecture $3,$ slide $25)$ applies?

Type of motion Direction of motion Type of acceleration

$+($speeding up$)++$ 

$–($slowing down$)–+$ 

$+($speeding up$)––$ 

$–($slowing down$)+–$Toronto

 

Transcribed Image Text:

Huntsville 342 km Ottawa Ꮎ 236 km Toronto N Huntsville 342 km Ottawa Ꮎ 236 km Toronto N

Answer rating: 100% (QA)

This problem involves concepts from physics such as displacement velocity acceleration and vectors a To calculate the magnitude of Chriss displacement z we can use the Pythagorean theorem because the ... View the full answer