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.

Respuesta :

Answer:

81.25 m

Explanation:

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

[tex]49*\frac {1000}{3600}=13.61 m/s[/tex]

From kinematics equation

[tex]v^{2}=u^{2}+2as[/tex] where v is final velocity, u is initial velocity, a is acceleration, s is distance and making a the subject of formula, [tex]a=\frac {v^{2}-u^{2}}{2s}[/tex] 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

[tex]a=\frac {0^{0}-13.61^{2}}{2*13}=-7.12547 m/s^{2}[/tex]

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

[tex]122.5*\frac {1000}{3600}=34.03 m/s[/tex]

From kinematics equation

[tex]v^{2}=u^{2}+2as[/tex]

making s the subject of formula

[tex]s=\frac {v^{2}-u^{2}}{2a}[/tex]

Substituting 0 for v, 34.03 m/s for u, [tex]-7.12547 m/s^{2}[/tex] for a we obtain

[tex]s=\frac {0^{0}-34.03^{2}}{2*-7.12547}=81.25 m[/tex]