A car moving along a straight highway with speed of 126 km h^(-1)is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop ?

A car moving along a straight highway with speed of 126 km h^(-1)is br

1.	A car moving along a straight highway with speed of 126 km h^(-1)is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop ?
Answer» Solution :Initial velocity of CAR, `u=126 kmh^(-1) =126 xx (5)/(18)ms^(-1)=35ms^(-1)"" ...(i)` Since, the car finally comes to rest, `v=0` Distance COVERED, `s=200m, a=?, t=?` ` v^(2)= u^(2)-2as` or ` a=(v^(2)-u^(2))/(2s)"" ...(ii)` Substituting the values from eq. (i) in eq. (ii), we get `a=(0-(35)^(2))/(2xx200)= -(35xx35)/(400)` `=- (49)/(16)ms^(-2)` `= -3.06 ms^(-2)` Negative sign shows that acceleration in negative which is called retardation, i.e., car is uniformly retarded at `-a=3.06 ms^(-2)`. To find t, let us USE the relation `v=u+ a t` `t=(v-u)/(a)` Use `a=-3.06ms^(-2), v=0, u=35ms^(-1)`. `thereforet=(v-u)/(a)=(0-35)/(-3.06)=11.44s` `therefore t=11.44 sec.`