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Question: Given, 3hrs 40min on a foggy day, two drivers spot in front of each other when 80m apart. They were ...

Given, 3hrs 40min on a foggy day, two drivers spot in front of each other when 80m apart. They were travelling at 70kmph and 60kmph both applied brakes simultaneously which retard the cars at the rate of 5m/s. Which of the following statements is correct?
a).The collision will be averted
b).The collision will take place
c).They will cross each other
d).They will just collide

Explanation

Solution

In this question, we will use the equation of motion which gives relation between the distance, final velocity, initial velocity and acceleration on an object. Further we will substitute the given values to find the required answer. Also, we will see the basics of the equation of motion and the law of conservation of momentum.
Formula used:
v2u2=2asv_{}^2 - u_{}^2 = 2a{s_{}}

Complete answer: :
Here, we first find the two initial velocities, given as:
uA=70×518=19.44m/s{u_A} = 70 \times \dfrac{5}{{18}} = 19.44m/s
uB=60×518=16.67m/s{u_B} = 60 \times \dfrac{5}{{18}} = 16.67m/s
Using the equation of motion to find the distance travelled:
vA2uA2=2asAv_A^2 - u_A^2 = 2a{s_A}
Now, we will use the equation of motion to find the distance travelled:
(19.44)2=2×5×sA- {(19.44)^2} = 2 \times 5 \times {s_A}
sA=37.9m\Rightarrow {s_A} = 37.9m
Similarly, we use the equation of motion to find the distance travelled by car B:
vB2uB2=2asBv_B^2 - u_B^2 = 2a{s_B}
Now, by substituting the given values, we get:
(16.67)2=2×5×sB- {(16.67)^2} = 2 \times 5 \times {s_B}
sB=27.78m\Rightarrow {s_B} = 27.78m
sA+sB<80\because {s_A} + {s_B} < 80

Therefore, the two cars will not crash and option a) is the correct answer i.e., the collision will be averted.

Additional information:
As we know that the equations of motion are the equations which describe the behavior of any physical system in terms of its motion as a function of time. Further we can also say that the equations of motion tell us the behavior of any physical system as a set of mathematical functions in terms of different dynamic variables. Here, these dynamic variables are said to be generally spatial coordinates and time are used, but other variables are also possible, for example- momentum components and time.
Also, from the law of conservation of mass, we know that the mass can neither be created nor destroyed in any chemical reactions.
We know that collision means when two objects come in contact with each other for a very short period. Here are two types of collision: first is elastic collision where the energy remains the same after interaction or collision and second is inelastic collision where the final energy changes after the collision of the particles or body.

Note:
Here, we should remember that the three different equations of motion are used in finding different physical properties of a particle under motion. We should also observe that these equations are only applicable to the classical system not in the quantum system. Further, during the solution, one should see that laws of conservation are not violated, i.e., mass can neither be created nor be destroyed and similarly for energy of the particles as well.