Question

6. A scooter moving with a velocity of 18 km/h is brought rest in 5sec by applying breaks.What is the
retardation produced?

Answer 1

its an easy one
according to equation of motion
v=u-at
at=u-v
a=(18-0)÷5
r=18÷5

hence retardation =3.6ms^(-2)

Answer 2

Explanation :

Initial velocity of the scooter, u = 18 km/h = 5 m/s

It is brought to rest i.e final velocity, v = 0

Time taken , t = 5 s

Retardation is same as deceleration or negative acceleration and it can be calculated as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{0-5\ m/s}{5\ s}$

$a=-1\ m/s^2$

Hence, the retardation produced is $-1\ m/s^2$.