Determine the maximum theoretical speed that may be...

Determine the maximum theoretical speed that may be achieved over a distance of $60 \mathrm{m}$ by a car starting from rest, knowing that the coefficient of static friction is 0.80 between the tires and the pavement and that 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) four-wheel drive, $(b)$ front-wheel drive, $(c)$ rear-wheel drive.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 So here you have to determine the maximum critical speed that must be achieved over a distance of 60 meters by a car starting from rest, knowing that the coefficient of static friction is 0 .8 between the tires and the pavement and 60 % of the weight of the car is distributed over its front wheel and 40 % over its rear wheel.

00:26 Now a, you have to assume a 4 -wheel rife, right? so this is pretty simple we start with a which is a four -wheel drive car right so we have okay so what happens is we have the normal reactions of from the tires that is acting upwards so this is let's say n1 and n2 and we will have the frictional force is f1 and f2 in the wheels right so let's say this is the wheel and this the another wheel right so this is the forces that are acting now we have the summation of forces along the y axis should be equal to zero and the weight is acting downwards so this we have n1 plus n2 this is equal to the weight w right so this is a equation and next we have the frictional forces f1 plus f2 that is equal to mu times of n1 plus new times of n2 right or this is equal to mu times of n1 plus n2 which is equal to w right now the sum of this forces friction forces must be equal to m a right which with its acceleration so a will be equal to mu g because weight is equal to m a and m and m will get cancelled you will have g right now if you plug in the values this is equal to 0 .8 times of of 9 .8 meters per second square which is equal to 7 .85 we can say meters per second square right next to find out the final velocities so v square equal to u square plus 2 as is a kinematic equation and here we have v square equal to u square is zero because it started from to as and this is equal to two times of a is 7 .85 and s we already have a 60 meters it's stopping distance sorry you have to this is the distance covered right and this will be we will be equal to then 30 .69 meters per second right right so this is the solution to this first question...

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