Question

Two cars A and B are approaching each other
head-on with speeds 20 m/s and 10 m/s
respectively. When their separation is x then A and
B start braking at 4 m/s2 and 2m/s2 respectively.
Minimum value of X to avoid collision is
(1) 60 m
(2) 75 m
(3) 80 m
(4) 90 m​

Answer 1

● Answer -

(2) X = 75 m

◆ Explanation -

Let s1 & s2 be stopping distances for car A & car B respectively.

Applying Newton's 3rd law of kinematics to car A,

v1² = u1² + 2as

0 = 20² + 2×(-4) × s

s1 = 400 / 8

s1 = 50 m

Applying Newton's 3rd law of kinematics to car B,

v2² = u2² + 2as

0 = 10² + 2×(-2) × s

s2 = 100 / 4

s2 = 25 m

Therefore, minimum separation between cars at the time of breaking is -

X = s1 + s2

X = 50 + 25

X = 75 m

Hope I was useful.❤️❤️❤️

Answer 2

• Let S is the distance to stop the car A.

and

• S' is the distance to stop the car B.

___________ [ ASSUMPTION ]

• Initial velocity for car A = 20 m/s

• Initial velocity for car B = 10 m/s

• Final velocity for car A = 0 m/s

• Final velocity for car B = 0 m/s

• Acceleration for car A = - 4 m/s²

• Acceleration for car B = - 2 m/s²

_______________ [ GIVEN ]

» We know that

v² - u² = 2as

(From Third Law of Motion .. By - Newton)

Put the know values in above formula

» CASE 1. (for car A) :-

=> (0)² - (20)² = 2(-4)S

=> - 400 = - 8S

=> 8S = 400

=> S = 50 m ________ (eq 1)

______________________________

» CASE 2. (for car B)

=> (0)² - (10)² = 2(-2)S'

=> - 100 = - 4S'

=> 4S' = 100

=> S' = 25 m __________ (eq 2)

______________________________

Minimum value of X to avoid collision is

=> S + S'

=> 50 + 25

=> $\textbf{75\:m}$ [Option 2)]

_____________________________