A car is traveling at 51.8 km/h on a flat highway.
(a) If the coefficient of friction between the road and the tires on a rainy day is 0.183, what is the minimum distance in meteres needed for the car to stop?
+33665 The deceleration force, F, is F = 0.183*mass*g (g is acceleration of gravity). This equals -mass*a, where 'a' is acceleration (deceleration) of the car, so
a = -0.183*9.81m/s2. (The negative sign just means it is slowing down.)
With constant acceleration we have the kinetic equation v2 = u2 +2as, where v = final velocity (0), u = initial velocity (51.8km/hr = 51.8*103/3600 m/s) and s is distance travelled. So:
0 = (51.8*103/3600)2 - 2*0.183*9.81*s
s = (51.8*103/3600)2/2*0.183*9.81
$${\mathtt{s}} = {\frac{{\left({\frac{{\mathtt{51.8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{3}}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.183}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{57.663\: \!954\: \!885\: \!109\: \!463\: \!3}}$$ metres
or s ≈ 57.7 metres
+33665 The deceleration force, F, is F = 0.183*mass*g (g is acceleration of gravity). This equals -mass*a, where 'a' is acceleration (deceleration) of the car, so
a = -0.183*9.81m/s2. (The negative sign just means it is slowing down.)
With constant acceleration we have the kinetic equation v2 = u2 +2as, where v = final velocity (0), u = initial velocity (51.8km/hr = 51.8*103/3600 m/s) and s is distance travelled. So:
0 = (51.8*103/3600)2 - 2*0.183*9.81*s
s = (51.8*103/3600)2/2*0.183*9.81
$${\mathtt{s}} = {\frac{{\left({\frac{{\mathtt{51.8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{3}}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{0.183}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{57.663\: \!954\: \!885\: \!109\: \!463\: \!3}}$$ metres
or s ≈ 57.7 metres
