Question

A body traveling at a velocity of 30m/s in a straight line is brought to rest by the application Of breaks. If it covers a distance of 100m during this period, find the retardation(please solve it in the simplest way)

Answer 1

initial velocity(u)- 30m/s

final velocity(v)- 0

distance =100m

formula-v^2=u^2+2as

0=30^2+2a*100

0=900+200a

200a = -900

a=-900/200

a=-9/2ms^-2

retardation=9/2m/s

hope it helps you

Answer 2

Answer:

I think the correct answer is given above

u equals initial velocity

v equals final velocity

and s equals distance.

a equals acceleration

BUT SINCE a HAS NEGATIVE VALUE THEREFORE IT DENOTES RETARDATION.

SO RETARDATION IS 4.5m/s(WITHOUT NEGATIVE SIGN).

HOPE IT HELPS