Question

an object is moving with a velocity of 6 metre per second and acceleration of -1 metre per second^2 find the distance covered before it come to rest.​

Answer 1

Answer: 18m

Explanation:

V^2 - U^2 = 2as

V is final velocity which must be 0 for object to come in rest position.

U is initial velocity by which the object is moving which is 6m/s.

a is acceleration which is -1 m^2/s that is deacceleration because acceleration is in negative.

And s is the displacement.

0^2 - (6)^2 = 2*(-1)*s

0 - 36 = -2s

s = 18m.

So it will take 18m to come to rest position.

Answer 2

Answer:

distance covered = 18 m

Explanation:

$v^{2} - u^{2} = 2 a s$

$s = \frac{v^{2} - u^{2}}{2 a}$

  $= \frac{0^{2} - 6^{2}}{2 (-1)}$

s = 18 m

                                      OR

v = u + at => t = (v - u) / a = (0 - 6) / -1 = 6 s

$s = ut + \frac{1}{2} a t^{2}$

  = 6 x 6 + (1/2) (-1) 6^2

  = 36 - 18 = 18 m

s = 18 m