Question

two cars A and B are approaching Each Other head on with speed 20 metre per second and 10 metre per second respectively their separation is X then A and B start breaking at 4 metre per second square into metre per second square respectively minimum value of X to avoid collision is​

Answer 1

Dear Students,

● Answer -

X = 62.5 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×(-4) × s

s2 = 100 / 8

s2 = 12.5 m

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

X = s1 + s2

X = 50 + 12.5

X = 62.5 m

Hope I was useful..

Answer 2

Answer:

up

Explanation:

refer to attachment ......