# Question #c9885

Feb 7, 2017

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

#### Explanation:

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

From Newton's Second Law of motion
Net horizontal deceleration $= \text{Resistive Force"/"Mass}$
$\therefore \text{acceleration } a = - \frac{1.7 \times {10}^{4}}{1100} = - 15. \overline{45} m {s}^{-} 2$

Using the kinematic equation

${v}^{2} - {u}^{2} = 2 a s$
where $v$ is final velocity, $s$ is displacement

Inserting various values we get
${v}^{2} - {24}^{2} = 2 \times \left(- 15. \overline{45}\right) \times 18$
$\implies {v}^{2} = {24}^{2} - 556. \overline{36}$
$\implies {v}^{2} = 19. \overline{63}$
$\implies v = \sqrt{19. \overline{63}}$
$\implies v = 4.43 m {s}^{-} 1$, rounded to two decimal places.