Question

Please answer with clear explanation . (NO SPAMMING )​

Answer 1

Answer:

$<p style="color:cyan;font-family:cursive;background:black;font size:40px;"><strong><em><u>The car will move 13.60 m</u></em></strong><strong><em><u> </u></em></strong><strong><em><u>Before the driver reacts, the car will have moved</u></em></strong></p><p><strong><em><u>Before the driver reacts, the car will have moveds1=8.33⋅0.7m=5.831m The time for the car to stop can be found fromv</u></em></strong></p><p><strong><em><u>Before the driver reacts, the car will have moveds1=8.33⋅0.7m=5.831m The time for the car to stop can be found fromv1=v0+at</u></em></strong><strong><em><u>0=9.33−5t or t=1.866 s</u></em></strong></p><p><strong><em><u>0=9.33−5t or t=1.866 sThe average speed during this time is </u></em></strong></p><p><strong><em><u>0=9.33−5t or t=1.866 sThe average speed during this time is va=12⋅8.33m</u></em></strong><strong><em><u>s</u></em></strong><strong><em><u>=4.165m</u></em></strong></p><p><strong><em><u>d</u></em></strong><strong><em><u>Therefore, the car moves </u></em></strong></p><p><strong><em><u>Therefore, the car moves 4.165</u></em></strong></p><p><strong><em><u>Therefore, the car moves 4.165⋅1.866</u></em></strong></p><p><strong><em><u>Therefore, the car moves 4.165⋅1.866 m =7.772m</u></em></strong></p><p><strong><em><u>Therefore, the car moves 4.165⋅1.866 m =7.772mTotal distance moved before the car stops, therefore, is</u></em></strong></p><p><strong><em><u>Therefore, the car moves 4.165⋅1.866 m =7.772mTotal distance moved before the car stops, therefore, is(5.831+7.772) m = 13.603 m</u></em></strong></p><p><strong><em><u>Therefore, the car moves 4.165⋅1.866 m =7.772mTotal distance moved before the car stops, therefore, is(5.831+7.772) m = 13.603 mRounded to 2 decimals: 13.60 m</u></em></strong><strong><em><u>.</u></em></strong><strong><em><u> </u></em></strong><strong><em><u></p>$

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Answer 2

$\LARGE\mathtt{Answer:}$

$\huge\underline{\underline\bold{12.77\;m}}$

$\large\mathtt{Explanation:}$

$a\;=\;-5 m/s²$

$u\;=\;8.33 m/s$

$v\;=\;0$

${v}^{2}\;=\;{u}^{2}\;+\;2as$

$0\;=\;(8.33{)}^{2}\;–\;105$

$⇒\; \frac{(8.33{)}^{2}}{10}=\;s\;=\;6.93$

$Distance\;in\;react\;time$

$=\;0.7\;×\;8.33$

$=\;5.831$

$Total\;=\;12.769$

$Therefore,$$\huge\underline{\underline\bold{12.77\;m}}$

$Hope\;this\;Helps:)$