Question

Part One:

Answer 1

Answer:

(a) $-7.95 m/s^{2}$

(b) 43.45 m

(c) 12.58 m

Explanation:

By applying newton's law on car,  

$F_{net}=ma$ where $F_{net}$is the friction force and m is mass, a is acceleration

$F_{net}=F_r = -\mu*m*g$

$g=9.8 m/s^{2}$

a = acceleration of car $a=\frac {f_r}{m}$

$a=-\mu*g=0.811*9.8$

$a=-7.9478 m/s^{2}\approx -7.95 m/s^{2}$

(b)  

From kinematic equation  

$v^{2}=u^{2}+2as$ but u=0 hence  

$v^{2}=2as$ and making s the subject  

$s=\frac {v^{2}}{2a}$  

$s=\frac {-26.3^{2}}{2*-7.95}=43.50252\approx 43.50 m$  

(c  

From the kinematic equation

$s=ut+0.5at^{2}$

$s=26.30.519+0.5(-7.95*0.519^{2})=13.6497-1.07071=12.57899\approx 12.58 m$