Question 101: Motion in 1D

Question

Speeds of two identical cars are $u $ and $4$$u $ at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is :

Accepted Answer

Given the initial speeds of two identical cars as $u$ and $4u$, and considering that both cars eventually stop (final speed $v = 0$), we note that both cars decelerate with the same acceleration $-a$.
Using the kinematic equation:
$ v^2 = u^2 - 2as $
Since $v = 0$,
$ 0 = u^2 - 2as $
Hence,
$ u^2 = 2as $
For the first car with speed $u$,
$ u^2 = 2a s_1 \quad 	ext{...(i)} $
For the second car with speed $4u$,
$ (4u)^2 = 2a s_2 \quad 	ext{...(ii)} $
Dividing equation (i) by equation (ii),
$ \frac{u^2}{(4u)^2} = \frac{2a s_1}{2a s_2} $
$ \frac{u^2}{16u^2} = \frac{s_1}{s_2} $
$ \frac{1}{16} = \frac{s_1}{s_2} $
Thus, the ratio of the stopping distances of the two cars is $\frac{1}{16}$.

Answer

$1:1$

Answer

$1:4$

Answer

$1:8$

Answer

$1:16$

Motion in 1D

Speeds of two identical cars are u and 4u at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is :

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