Question

A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of friction between road and tires on a rainy day is 0.096, what is the minimum distance in which the car will stop? (1 mi = 1.609 km) Answer in units of m.

Answer 1

The minimum distance in which the car will stop is

x=167.38m

Step-by-step explanation:

`39.7(mi)/(h)*(1km)/(0.621371mi)*(1000m)/(1km)*(1h)/(3600s)=17.747(m)/(s)`

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

`F=m*a\\a=(F)/(m)\\a=(u*m*g)/(m)\\a=u*g\\a=0.096*-9.8(m)/(s^(2) )`

`a=-0.9408 (m)/(s^(2))`

`v_(f)^(2)=v_(o)^(2)+2*a*(x_(f)-x_(o))\\v_(f)=0 \\x_(o)=0\\0=v_(o)^(2)+2*a*x_(f)\\x_(f)=(v_(o)^(2))/(2*a) \\x_(f)=((-17.747(m)/(s))^(2))/(2*(-0.9408)) \\x_(f)=167.38m`