Question

This question is typical on some driver's li-

cense exams: A car moving at 49 km/h skids
17 m with locked brakes.
How far will the car skid with locked brakes
at 98 km/h? Assume that energy loss is due
only to sliding friction.
Answer in units of m.

Answer 1

`d=68m`

Step-by-step explanation:

Initial | Final

`v^(2) -v_(o) ^(2) =2ad` |
`v^(2) -v_(o) ^(2) =2ad`

`0^(2) -49^2=2a(17)` |
`0^(2) -98^2=2ad`

`(49^2)/(98^2) =(17)/(d)`

`d=68m`

Answer 2

The car will skid 98 m

Step-by-step explanation:

Friction Loss

If some object is traveling and the brakes are locked, then the friction force dissipates all the kinetic energy into heat.

The kinetic energy dissipated is equal to the job done by the friction force:

`K = W_f\qquad\qquad[1]`

The kinetic energy is calculated by:

`\displaystyle K=(1)/(2)mv^2`

Where m is the object's mass and v is the speed.

The work done by the friction force is:

`W=F_r.x`

Where Fr is the magnitude of the force and x is the distance the object is braking.

The friction force is calculated as

`F_r=\mu .m.g`

Being
`\mu` the coefficient of friction and g the acceleration of gravity. Please note that the friction force remains constant for a given object and surface of contact.

Keeping all the above in mind, we use [1]

`\displaystyle (1)/(2)mv^2=\mu .m.g.x`

Simplifying and solving for x:

`\displaystyle x=(v^2)/(2\mu g)`

Which means the skidding distance is proportional to the square of the speed. Or, equivalently for x2 and x1 the distances for v1 and v2:

`\displaystyle (x_2)/(x_1)=\left((v_2)/(v_1)\right)^2`

If the car skids 17 m at 49 Km/h, then if it goes at 98 Km/h, it will skid:

`\displaystyle x_2=17*\left((98)/(49)\right)^2=68`

`x_2=68~m`

The car will skid 98 m