Question

please quickly answer please​

Answer 1

Explanation:

Given:

Accerleration: a=−6 m/s

2

Time t=2 s

Final velocity, v=0 m/s

v=u+at

0=u−6×2

u=12 m/s

s=ut+ 21at^2

s=12×2+ 21 ×(−6)×4

⇒s=12 m

Answer 2

Answer:

Kinematics

Explanation:

★ Given ;

» Retardation ( a ) = - 6 m / s²

[ negative bcoz it acts opposite to original motion ]

» Stopping Time ( t ) = 12 seconds

» Final Velocity ( v ) = 0 ( as it stops )

★ Initial velocity ( U ) = ???

★ Stopping distance ( s ) = ???

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★ From equation of Motion ;

★ V = U + a t

0 = U - a t

U = a × t

= 6 × 12

★ Initial Velocity ( u ) = 72 m / s ★

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★ From equation of Motion;

V² - U² = 2 × a × s

S = U² / 2 × a

= 72 × 72 / 2 × 6

= 72 × 72 / 12

= 6 × 72

★ Stopping Distance ( s ) = 432 metres ★

Hope it helps you .