Question

a body is accelerated from rest to a velocity of 50 m per sec over a dustance of 0.250 m find the acceleration of the body​

Answer 1

Answer:

5000 m/s^2

Explanation:

Using the third equation of motion

$v^2=u^2+2as\\ 50^2=0^2+2a*0.250\\a=5000 m/s^2$

Answer 2

Given :

• Initial velocity of body, u = 0   [since body is at rest]
• Final velocity of body, v = 50 m/s
• Distance covered by body while accelerating. s = 0.250 m

To find :

• Acceleration of the body, a = ?

Formula required :

Third equation of motion

• 2 a s = v² - u²

[ where a is acceleration, s is distance covered, v is final velocity and u is initial velocity ]

Calculation :

Using third equation of motion

→ 2 a s = v² - u²

→ 2 a (0.250) = (50)² - (0)²

→ 0.5 a = 2500

→ a = 2500 / 0.5

→ a = 25000 / 5

→ a = 5000 m/s²

therefore,

• Acceleration of the body is 5000 m/s² .

Three equation of motion are as follows :

• First equation of motion

       v = u + a t

• Second equation of motion

    s = u t + 1/2 a t²

• Third equation of motion

     2 a s = v² - u²

[ where a is acceleration, s is distance covered, v is final velocity and u is initial velocity and t is time taken ]