Question #c9885

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Question #c9885

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Answer 1 · 2017-02-07T23:36:30

$4.43 m s^{- 1}$ $4.43ms^-1$ , rounded to two decimal places.

Explanation:

Given initial speed $u$ $u$ of car before it enters patch of mud $= 24 m s^{- 1}$ $=24ms^-1$ .
Net horizontal resistive force in mud patch $= 1.7 \times 10^{4} N$ $=1.7xx10^4N$

From Newton's Second Law of motion
Net horizontal deceleration $= \frac{Resistive Force}{Mass}$ $="Resistive Force"/"Mass"$
$:."acceleration "a=-(1.7xx10^4)/1100=-15.bar45ms^-2$

Using the kinematic equation

$v^2-u^2=2as$
where $v$ is final velocity, $s$ is displacement

Inserting various values we get
$v^2-24^2=2xx(-15.bar45)xx18$
$=>v^2=24^2-556.bar36$
$=>v^2=19.bar63$
$=>v=sqrt(19.bar63)$
$=>v=4.43ms^-1$ , rounded to two decimal places.

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Question #c9885

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