Physics, asked by Anonymous, 7 months ago

The brakes applied to a car produce an acceleration of 6ms -2 in the opposite direction to the motion. If

the car takes 2s to stop after the application of brakes, calculate the difference it travels during this time

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Answers

Answered by Anonymous

Question :

The brakes applied to a car produce an acceleration of 6 m/s² in the opposite direction to the motion. If the car takes 2s to stop after applying brakes, calculate the distance it travels during this time .

Solution :

Final velocity (v) = 0 m/s
Time interval (t) = 2 s
Acceleration (a) = - 6 m/s²

Acceleration is Taken as negative because it is in opposite direction of the motion. We will use Kinematics equations.

$\underbrace{\sf{Initial \: Velocity}}$

Use 1st equation of Kinematics

$\implies \sf{v \: = \: u \: + \: at} \\ \\ \implies \sf{v \: - \: u \: = \: at} \\ \\ \implies \sf{0 \: - \: u \: = \: 2 \: \times \: -6} \\ \\ \implies \sf{-u \: = \: -12} \\ \\ \implies \sf{u \: = \: 12} \\ \\ \underline{\sf{\therefore \: Initial \: velocity \: is \: 12 \: ms^{-1}}}$

________________________________

$\underbrace{\sf{Distance \: travelled \: by \: car}}$

Use 3rd equation of Kinematics

$\implies \sf{v^2 \: - \: u^2 \: = \: 2as} \\ \\ \implies \sf{s \: = \: \dfrac{v^2 \: - \: u^2}{2a}} \\ \\ \implies \sf{s \: = \: \dfrac{0^2 \: - \: 12^2}{-12}} \\ \\ \implies \sf{s \: = \: \dfrac{-144}{-12}} \\ \\ \implies \sf{s \: = \: 12} \\ \\ \underline{\sf{\therefore \: Distance \: travelled \: by \: car \: is \: 12 \: m}}$