Question

001 10.0 points

This question is typical on some driver's li-
cense exams: A car moving at 49 km/h skids
13 m with locked brakes.
How far will the car skid with locked brakes
at 122.5 km/h? Assume that energy loss is
due only to sliding friction.
Answer in units of m.

Answer 1

81.25 m

Step-by-step explanation:

49 Km/h is initial velocity, converted to m/s we get

`49*\frac {1000}{3600}=13.61 m/s`

From kinematics equation

`v^(2)=u^(2)+2as` where v is final velocity, u is initial velocity, a is acceleration, s is distance and making a the subject of formula,
`a=\frac {v^(2)-u^(2)}{2s}` and when it skids, final velocity is zero hence substituting 0 m/s for v, 13.61 m/s for u and 13 m for s

`a=\frac {0^(0)-13.61^(2)}{2*13}=-7.12547 m/s^(2)`

When the initial velocity is 122.5 Km/h, converted to m/s

`122.5*\frac {1000}{3600}=34.03 m/s`

From kinematics equation

`v^(2)=u^(2)+2as`

making s the subject of formula

`s=\frac {v^(2)-u^(2)}{2a}`

Substituting 0 for v, 34.03 m/s for u,
`-7.12547 m/s^(2)` for a we obtain

`s=\frac {0^(0)-34.03^(2)}{2*-7.12547}=81.25 m`