Question

Two cars 1 and 2 move with velocities ‘\[{{v}_{1}}\]’ and ‘\[{{v}_{2}}\]’ respectively, on a straight road in same direction. When the cars are separated by a distance ‘d’, the driver of car 1 applies brakes and the car moves with uniform retardation ‘${{a}_{1}}$’. Simultaneously, 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.

Answer 1

Hint: We have two cars moving with different velocity in the same direction on a straight road. By using the third kinematic equation we can formulate the relation of relative motion. Using this equation and the condition to avoid collision we will get the solution.
Formula used:
${{v}^{2}}={{u}^{2}}+2as$

Complete answer:
In the question it is said that two cars, car 1 and car 2 are moving in the same direction on a straight road with different velocities.
The initial velocity of the car 1 is ‘\[{{v}_{1}}\]’ and of car 2 is ‘\[{{v}_{2}}\]’.
It is given that \[{{v}_{1}}>{{v}_{2}}\]
The distance between the two cars is given as ‘d’.
The retardation of car 1 when the driver applies break is ‘${{a}_{1}}$’ and the acceleration of car 2 is ‘${{a}_{2}}$’.
From the third equation of motion, the relative motion between the two cars can be written as,
\[v_{rel}^{2}=u_{rel}^{2}+2{{a}_{rel}}{{s}_{rel}}\], were ‘v’ is final velocity, ‘u’ is initial velocity, ‘a’ is acceleration and ‘s’ is the separation.
Let ‘\[v_{1}^{'}\]’ be the final velocity of the car 1 and ‘\[v_{2}^{'}\]’ be the final velocity of car 2.
Then the relative final velocity can be written as,
${{v}_{rel}}=v_{1}^{'}-v_{2}^{'}$
And the relative initial velocity can be written as,
${{u}_{rel}}={{v}_{1}}-{{v}_{2}}$
Relative acceleration will be,
${{a}_{rel}}=-{{a}_{1}}-{{a}_{2}}$
$\Rightarrow {{a}_{rel}}=-\left( {{a}_{1}}+{{a}_{2}} \right)$
And the relative separation,
${{s}_{rel}}=-d$
By substituting this in these values in the relative equation, we get
$\Rightarrow {{\left( v_{1}^{'}-v_{2}^{'} \right)}^{2}}={{\left( {{v}_{1}}-{{v}_{2}} \right)}^{2}}+2\left( -\left( {{a}_{1}}+{{a}_{2}} \right) \right)\left( -d \right)$
$\Rightarrow {{\left( v_{1}^{'}-v_{2}^{'} \right)}^{2}}={{\left( {{v}_{1}}-{{v}_{2}} \right)}^{2}}+2\left( {{a}_{1}}+{{a}_{2}} \right)\left( d \right)$
In the question we are asked to find the minimum initial separation to avoid collision.
We know that to avoid collision the final velocity of car 1 should be less than the final velocity of car 2, i.e. $\left( \mathop{v}_{1}^{'}-\mathop{v}_{2}^{'} \right)$
By applying this condition in the equation, we get the minimum distance as,
$\Rightarrow d=\dfrac{{{\left( {{v}_{1}}-{{v}_{2}} \right)}^{2}}-{{\left( v_{1}^{'}-v_{2}^{'} \right)}^{2}}}{2\left( {{a}_{1}}+{{a}_{2}} \right)}$
This is the minimum distance to avoid collision.

Note:
Relative motion of a body is the motion of that body with respect to some other, adjacent frame of reference. Every motion will always be relative to some frame of reference.
If there are two objects A and B, the relative motion of the object A with respect to B is said as the rate of change of position of A with respect to the object B.