Question

An object is rolled at 12 m/s dowb a table. It stops after 15 s. What was it's acclaration ?​

Answer 1

Explanation:

here,

the beginning velocity,u=12m/s

last velocity, v=0m/s

time,t=15s

we know that,

acclaration,a=v-u/t=0-12/15=-0.8m/s^2

hope its help.Thanks.

Answer 2

Correct Question :-

An object is rolled at 12 m/s down a table. It stops after 15 s. What was it's acclaration ?

Given :-

• An object is rolled at 12 m/s down a table.
• It stops after 15 s.

To Find :-

• Acceleration

Solution :-

We have,

• Final Velocity(v) = 0 m/s
• Initial Velocity(u) = 12 m/s
• Time taken(t) = 15 s

We know that,

$\boxed{\sf v = u + at} \: \: [ \bf 1st \: equation \: of \: motion ]$

Substituting the given values we get,

➠ 0 = 12 + a(15)

➠ 0 = 12 + 15a

➠ - 15a = 12

➠ a = 12/(- 15)

➠ a = (- 4)/5 $\sf m/s^{2}$

∴ The acclaration is (- 4)/5 $\bf m/s^{2}$

_______________________________

Additional Information :-

☆ 2nd equation of motion :-

$\boxed{\sf s = ut + \dfrac{1}{2}at^{2}}$

Where,

• s = Distance covered
• u = Initial Velocity
• t = Time taken
• a = Acceleration

☆ 3rd equation of motion :-

$\boxed{\sf v^{2} - u^{2} = 2as}$

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance covered

_______________________________