Question 13 Marks
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
View full question & answer→Initial velocity of car,
$u =126 kmh ^{-1}=126 \times \frac{5}{18} ms ^{-1}=35 ms ^{-1}$
Since, the car finally comes to rest, $v=0$
Distance covered, $s =200 m , a =?, t =$ ?
$v ^2= u ^2-2 as$
or $a =\frac{v^2-u^2}{2 s}$
substituting the values from eq. (i) in eq . (ii), we get
$a =\frac{0-(35)^2}{2 \times 200}=\frac{0-(35)^2}{2 \times 200}$
$=-\frac{46}{16} ms ^{-2}=-3.06 ms ^{-2}$
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
$v = u + at$
$t =\frac{v-u}{a}$
use $a =-3.06 ms ^{-2}, v =0, u =35 ms ^{-1}$
$\therefore t =\frac{v-u}{a}=\frac{0-35}{-3.06}=11.44 s$
$\therefore t =11.44 sec$
$u =126 kmh ^{-1}=126 \times \frac{5}{18} ms ^{-1}=35 ms ^{-1}$
Since, the car finally comes to rest, $v=0$
Distance covered, $s =200 m , a =?, t =$ ?
$v ^2= u ^2-2 as$
or $a =\frac{v^2-u^2}{2 s}$
substituting the values from eq. (i) in eq . (ii), we get
$a =\frac{0-(35)^2}{2 \times 200}=\frac{0-(35)^2}{2 \times 200}$
$=-\frac{46}{16} ms ^{-2}=-3.06 ms ^{-2}$
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
$v = u + at$
$t =\frac{v-u}{a}$
use $a =-3.06 ms ^{-2}, v =0, u =35 ms ^{-1}$
$\therefore t =\frac{v-u}{a}=\frac{0-35}{-3.06}=11.44 s$
$\therefore t =11.44 sec$
