Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by a distance of 5 km, the speed of a car moving in the opposite direction if it meets these two cars at an interval of 4 minutes, will be

(1) 40 km/hr

(2) 45 km/hr

(3) 30 km/hr

(4) 15 km/hr

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Concept Questions :-

Relative motion

(2) The two car (say A and B) are moving with same velocity, the relative velocity of one (say B) with respect to the other $A,\text{​\hspace{0.17em}}{\stackrel{\to }{v}}_{BA}={\stackrel{\to }{v}}_{B}-{\stackrel{\to }{v}}_{A}=v-v=0$

So the relative separation between them (= 5 km) always remains the same.

Now if the velocity of car (say C) moving in opposite direction to A and B, is ${\stackrel{\to }{v}}_{C}$ relative to ground then the velocity of car C relative to A and B will be ${\stackrel{\to }{v}}_{rel.}={\stackrel{\to }{v}}_{C}-\stackrel{\to }{v}$

But as $\stackrel{\to }{v}$ is opposite to vC

So $\text{\hspace{0.17em}}{v}_{rel}={v}_{c}-\left(-30\right)=\left({v}_{C}+30\right)\text{\hspace{0.17em}}km\text{​​​}/\text{​​}hr.$

So, the time taken by it to cross the cars A and B $t=\frac{d}{{v}_{rel}}\text{​\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{4}{60}=\frac{5}{{v}_{C}+30}\text{\hspace{0.17em}\hspace{0.17em}}$

$⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{v}_{C}=45\text{\hspace{0.17em}}km/hr.$

Difficulty Level:

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