Question

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

Answer 1

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
• Intial Velocity(u) = 12 m/s
• Time taken(t) = 15 s

We know that,

→ a = (v - u)/t

Where,

• a = Acceleration
• v = Final Velocity
• u = Initial Velocity
• t = Time taken

Substituting the values we get,

$\implies$a = (0 - 12)/15

$\implies$a = - 12/15

$\implies$a = - 4/5 m/s²

∴ The acceleration is - 4/5 m/s²

Answer 2

Answer:

We are given:

initial velocity (u) = 12 m/s

final velocity (v) = 0 m/s

time taken (t) = 15 seconds

acceleration (a) = a m/s²

Solving for acceleration:

from the first equation of motion

v = u + at

replacing the variables

0 = 12 + (a)(15)

0 = 15a + 12

a = -12 / 15

a = -4 / 5 m/s²