Question

The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement, the friction of the road on the tires of the car, which is what slows down the car, is 25% of the weight of the car. What is the acceleration of the car? How many meters does it travel before stopping under these conditions?

Answer 1

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

$Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N$

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

$a=Fr/m$

$a=-4291.875/1750\\a=-2.4525$

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

$v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m$