The brakes applied to a car produce an accele | vedclass

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(A) Given: Acceleration $a = -6 \, m s^{-2}$ (negative because it is in the opposite direction of motion),time $t = 2 \, s$,and final velocity $v = 0

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$12$

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(A) Given: Acceleration $a = -6 \, m s^{-2}$ (negative because it is in the opposite direction of motion),time $t = 2 \, s$,and final velocity $v = 0 \, m s^{-1}$ (since the car stops).$(i)$ First,find the initial velocity $u$ using the first equation of motion: $v = u + at$.$0 = u + (-6) 	imes 2$$u = 12 \, m s^{-1}$.$(ii)$ Now,calculate the distance $S$ using the second equation of motion: $S = ut + \frac{1}{2}at^2$.$S = (12 	imes 2) + \frac{1}{2} 	imes (-6) 	imes (2)^2$$S = 24 - 12 = 12 \, m$.Therefore,the car travels $12 \, m$ before stopping.

The brakes applied to a car produce an acceleration of $6 \, m s^{-2}$ in the opposite direction of motion. If the car takes $2 \, s$ to stop after the application of the brakes,calculate the distance it travels during this time. (in $, m$)

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