Question

Two cars 1 and 2 move with velocities $v_{1}and{v}_2$ respectively on a straight road in the same direction. When the cars are separated by a distance d, the driver of the car 1 applies brakes and the car moves with uniform retardation $a_1$. Simultaneously, the car 2 starts accelerating with $a_2$. If $v_1>v_2$, find the minimum initial separation between the cars to avoid collision between them.

• A. $d=\frac{(v_1-v_2{)}^2}{\left(a_1{a}_2\right)}\frac{a_1}{a_2}$
• B. $d=\frac{2\left(a_1{a}_2\right)}{(v_1-v_2{)}^2}$
• C. $d=\frac{(v_1-v_2{)}^2(v_1)}{(a_1+a_2)v_2}$
• D. $d=\frac{(v_1-v_2{)}^2}{2(a_1+a_2)}$

Answer 1

The correct option is D

$d=\frac{(v_1-v_2{)}^2}{2(a_1+a_2)}$

Method 1
Let us assume that the cars undergo displacement $S_{1}and{s}_2$ after a time t from the instant or braking.
From figure, $s_1-s_2=d.....\left(i\right)$,
Where $s_1=v_{1}t-\frac{1}{2}{a}_1{t}_2....\left(ii\right)$
Using equations(i), (ii) and (iii), we have $(v_{1}t-\frac{1}{2}{a}_1{t}^2)-(v_{2}t-\frac{1}{2}{a}_1{t}^2)=d$
Or, $a_1+a_2)t^2-2(v_1-v_2)t+2d=0$
This is quadratic equation in t, solving for t
That gives t=$\frac{(v_1+v_2{)}^2-2(a_1+a_2)d}{a_1+a_2}$
If cars do not collide, $b^2-4ac<0\Rightarrow (v_1-v_2{)}^2-2(a_1+a_2)d<0$
$\frac{(v_1-v_2{)}^2}{2(a_1+a_2)}<dHence{d}_{min}=\frac{(v_1-v_2{)}^2}{2(a_1+a_2)}$

Method 2.
Relative velocity approach:
Using $v_{rel}^2=u_{rel}^2+2a_{rel}.s_{re}$ we can obtain the same result.
Substitute: $v_{rel}=v_{12}^{\prime }=v_1^{\prime }-v_2^{\prime },u_{rel}=v_1-v_2,$
$a_{rel}=a_{12}=(-a_1)-a_2=-(a_1+a_2)and{s}_{rel}=S_{12}=-d$, to obtain $(v_1^{\prime }-v_2^{\prime }{)}^2=(v_1-v_2{)}^2-2(a_1+a_2)d$
Since $v_1^{\prime }<v_2^{\prime }$ to avoid collision, we have d=$\frac{(v_1-v_2{)}^2-(v_1-v_2{)}^2}{2(a_1+a_2)}$
Maximum distance d for collision is possible is possible when $v_1^{\prime }=v_2^{\prime }$ then, $d_{max}=\frac{(v_1-v_2{)}^2}{2(a_1+a_2)}$