Two cars $ A $ and $ B $ are moving with same speed of $ 45\, km/h $ along same direction. If a third car $ C $ coming from the opposite direction with a speed of $ 36 \,km/h $ meets two cars in an interval of $ 5\, min $ , the distance of separation of two cars $ A $ and $ B $ should be (in km)
- 6.75
- 7.25
- 5.55
- 8.35
The Correct Option is A
Approach Solution - 1
$ {{s}_{rel}}={{v}_{rel}}\times t=81\times \frac{5}{18}\times (5\times 60) $
$ =\frac{81\times 5\times 5\times 60}{18}=6750m$
$=6.75\,km $
So, the correct option is (A) : 6.75.
Approach Solution -2
Given :
VA = VB = 45 km/hr
VC = 36 km/hr
Now, let's assume the following :
x be the distance between A and B.
Velocity of approach between A and B.
⇒ VAC = VA + VB
= (45 + 36) km/hr = 81km/hr
VBC = VB + VC
=(45 + 36) km/hr = 81km/hr
Now, Velocity of approach :
\(=\frac{\text{Distance between A and B}}{\text{Time difference in meeting A and B}}\)
\(⇒\Delta t=5\ \text{min}=\frac{5}{60}\)
\(⇒81=\frac{x\times60}{5}\)
\(⇒x=\frac{5\times81}{60}=\frac{405}{60}\)
\(=6.75\) km
So, the correct answer is (A) : 6.75
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Concepts Used:
Motion in a straight line
The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion.
Types of Linear Motion:
Linear motion is also known as the Rectilinear Motion which are of two types:
- Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
- Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.