Travelling at 18ms^-1 brakes are applied & car skids in a straight line to stop. Coefficient of friction is 0.4. How far does car travel before stopping?
OK, so,
F = μ N but N = mg
so F = μmg
& F = ma
=> μmg = ma
a = μg = 0.4 x 9.81 m s-2
a = 3.924 ms-2
The next bit should be easy, so why can't I do it?
Could you help find the answer please?
OK, so,
F = μ N but N = mg
so F = μmg
& F = ma
=> μmg = ma
a = μg = 0.4 x 9.81 m s-2
a = 3.924 ms-2
The next bit should be easy, so why can't I do it?
Could you help find the answer please?
-
g = 9.8 m/s^2
Coefficient of friction is 0.4
9.8 x 0.4 = friction acceleration = 3.92 m/s^2
0.5v^2 = a x distance
0.5 x 18^2 = 162
162/3.92 = 41.32 m
Coefficient of friction is 0.4
9.8 x 0.4 = friction acceleration = 3.92 m/s^2
0.5v^2 = a x distance
0.5 x 18^2 = 162
162/3.92 = 41.32 m