Question

A rule states that the frictional force between try concrete and a sliding car's tyre is equal to 9- 10th of the car's weight .if the skid marks left by a car in coming to rest are 20m long .how fast was the car going before the brakes were applied?

Answer 1

$\Huge\bold\red{Question}$

A rule states that the frictional force between try concrete and a sliding car's tyre is equal to 9- 10th of the car's weight .if the skid marks left by a car in coming to rest are 20m long .how fast was the car going before the brakes were applied?

$\Large\bold\green{Answer}$

$\sf \: normal \: force \: (n) = \: w \\ \\ \sf and \: friction \: (f) \: = \: 0.9 \: w \implies \: \: = \: 0.9 \\ \\ \sf \: also \: \: \: f \: = \: \frac{w}{g} a \\ \\ \sf \: 0.9 \: = \: \frac{w}{g} a \implies \: a \: = \: 0.9 \: g \: opposite \: in \: direction \: of \: motion \\ \\ \sf \: using \: \: {v}^{2} \: - {u}^{2} \: = \: 2as \implies \: u \: = \: \sqrt{ {v}^{2} - \: 2as} \\ \\ \sf \: = \: \sqrt{ {0}^{2} - 2( - 0.9g)(20) } \: = \: \sqrt{2 \times 0.9 \times 9.81 \times 20} \\ \\ \sf \: \sqrt{353.16} \: = \: 18.8 \: \frac{m}{s}$