Question

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

Answer 1

Step-by-step explanation:

Acceleration , a = -6m/s2

Time , t = 2sec

Final velocity , v = 0 m/s

By using 1st Equation of motion,

v=u+at

=>0=u+-6x2

=>u=12m/s

Initial velocity , u = 12m/s

By using 2rd equation of motion,

s=ut+1/2at^2

=>s=12×2+1/2×(-6)×2×2

=>s=24+(-12)

=>s=12m

Therefore distance covered is 12m

Hope this helps :)

Answer 2

Answer:

24 m

Step-by-step explanation:

U=6m/s

t=2sec

V=0(because of retardation)

a=?

S=?

To find a=

V=u-at

0=6-a×2

0=6-2a

-2a= -6(minus-minus cancelled out)

a=6/2

a=3

Now S,

S=ut+at^2

S=6×2+3×2×2

S=12+12

S=24m

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