A car moving along a straight highway with a speed of 126 kilometre per hour is brought to a stop within a distance of 200m. What is the retardation of the car (assumed uniform) and how long does it taken for the car to stop?

Explanation

Initial velocity of car,
`u=126kmh^(-1)=126xx(5)/(6)ms^(-1)=35ms^(-1)" "………(1)`
Since, the car finaly comes to rest, v=0
Distance covered, s 200 m, a=?, t=?
`v^(2)=u^(2)-2asora=(v^(2)-u^(2))/(2s)" "......(2)`
Substituting the values from eq. (1) in eq. (2), we get
`a=(0-(35)^(2))/(2xx200)=-(35xx35)/(400)=-(46)/(16)ms^(-2)=-3.06ms^(-2)`
Negative sign shows that acceleration in negative which is called retardation, uniformly retarded at `-a=3.06ms^(-2)`.
To find t, let us use the relation
v=u+at
`t=(v-u)/(a)`
Use `a=-3.06ms^(-2),v=0,u=35ms^(-1)`
`:.t=(v-u)/(a)=(0-35)/(-3.06)=11.44s`
`:.t=11.44` sec.