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

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Dear Student

$v^{2} - u^{2} = 2 a S a n d v = u + a t v = 0 u = - a t u = - 6 \times 2 = - 12 m / s e c (- s i g n i s b e c a u s e t h e d i r e c t i o n o f u a n d a i s o p p o s i t e) v^{2} - u^{2} = 2 a S 12 \times 12 = 2 \times 6 \times S S = 12 m R E G A R D S$

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

Dear Student v2-u2=2aSand v=u+atv=0u=-atu=-6×2=-12m/sec (-sign is because the direction of u and a is opposite)v2-u2=2aS12×12=2×6×SS=12mREGARDS

Dear Student

$v^{2} - u^{2} = 2 a S a n d v = u + a t v = 0 u = - a t u = - 6 \times 2 = - 12 m / s e c (- s i g n i s b e c a u s e t h e d i r e c t i o n o f u a n d a i s o p p o s i t e) v^{2} - u^{2} = 2 a S 12 \times 12 = 2 \times 6 \times S S = 12 m R E G A R D S$