Question

a body moving with uniform retardation covers 3km before its sped reduced to half of its initial value.it comes to rest is an9ther distance of

Answer 1

HEY BUDDY..!!!

HERE'S THE ANSWER..

______________________

▶️ We know

⏺️( v )^2 - ( u )^2 = 2 . a . s

▶️ Where ,

# u = initial velocity

# v = final velocity

# a = Acceleration

# s = displacement = 3 Km = 3000 m

✔️ In question it is given ( v = u / 2 )

⏺️ Putting all values in eqn

=> ( v )^2 - ( u )^2 = 2 . a . s

=> ( u / 2 )^2 - ( u )^2 = 2 . a . s

=> ( u )^2 / 4 - ( u )^2 = 2 . a . s

=> - 3 ( u )^2 / 4 = 2 . a . 3000

=> ( u )^2 = - 8000 . a __________( 1 )



▶️ Now , object comes to rest , so v = 0

=> ( v )^2 = ( u / 2 )^2 + 2 . a . s

=> - ( u )^2 / 4 = 2 . a . s

=> - ( u )^2 / 4 × 2 × a = s

=> - ( u )^2 / 8 × a = s

⏺️ Now putting value of ( s ) in eqn 1

=> s = - ( - 8000 . a ) / 8 . a

=> s = 1000 m = 1 Km



HOPE HELPED..!!!


JAI HIND !!


:-)