Question

On a two lane road, car A is travelling with a speed of \[36km{{h}^{-1}}\] . Two cars B and C approach car A in opposite directions with a speed of $54km{{h}^{-1}}$ each. At a certain instant, when the distance AB is equal to AC, both being $1km$ , B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
$A.9.8m{{s}^{-2}}$
$B.10m{{s}^{-2}}$
$C.1m{{s}^{-2}}$
$D.2.0m{{s}^{-2}}$

Answer 1

Hint: Concept of relative velocity will be used in order to solve the given problem. We will also use the second equation of motion in order to find the time duration of the motion. Relative velocity of a body is defined as the velocity with respect to another body.
Formula used:
We will use the following equation of motion to solve the given problem:-
$s=ut+\dfrac{1}{2}a{{t}^{2}}$.

Complete step by step solution:
From the problem given above we have following data with us:-
Speed of A, ${{v}_{A.}}=36km{{h}^{-1}}=36\times \dfrac{5}{18}=10m{{s}^{-1}}$
Speed of B, ${{v}_{B}}=54km{{h}^{-1}}=54\times \dfrac{5}{18}=15m{{s}^{-1}}$
Speed of C, ${{v}_{C}}=54km{{h}^{-1}}=54\times \dfrac{5}{18}=15m{{s}^{-1}}$.
Relative velocity of A with respect to C is given as
${{v}_{AC}}=10+15=25m{{s}^{-1}}$.
We have added the velocities to get the relative velocity as they are moving in opposite directions.
The final answer is in $m{{s}^{-1}}$therefore, we had converted the parameters in metre or $m{{s}^{-1}}$ .
Time taken by C to overtake A will be given as, $t=\dfrac{1000}{25}=40s$as distance between A and C remains $1km=1000m$.
Now distance travelled by A in this time duration is given as ${{s}_{A}}=40\times 10=400m$.
Now, B has to cover the distance of $1000+400=1400m$.to overtake A before C.
Now using second equation of motion we get,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$………………… $(i)$
Putting values of respective parameters in equation $(i)$ we get
$1400=15\times 40+\dfrac{1}{2}\times a\times {{40}^{2}}$
$\Rightarrow 1400=600+\dfrac{1}{2}\times a\times 1600$
$\Rightarrow 1400=600+800a$
$\Rightarrow 800a=1400-600$
$\Rightarrow 800a=800$
$\Rightarrow a=1m{{s}^{-2}}$

Hence, option $(C)$ is correct.

Note:
Concept of relative velocity should be used correctly. When two bodies approach each other in opposite directions then we add to find the relative velocity of one body with respect to another and when the bodies approach each other in the same direction then we subtract the respective velocities to find the relative velocity of one body with respect to another. Units should also be taken care of.