Question

On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B and
C approach car A in opposite directions with a speed of 54 km h-1 each. At a
certain instant, when the distance AB is equal to AC, both being 1 km, B decides
to overtake A before C does. What minimum acceleration of car B is required to
avoid an accident ?
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Answer 1

ANSWER

• Speed of A=36km/hr=36× 5/18 m/s
• =10m/s

Speed of B= Speed of C=54× 5/18 m/s

• =15m/s

Relative speed of A w.r.t C=10+15=25m/s

• Time taken by C to overtake A= 1000/25=40sec.

Distance travelled by A all this time =10×40=400m

• So, B have to cover distance of (1000m+400m)
• i.e. 1400m to take over a A before C does. in 40sec.

Now putting in formula;-

$s=ut+ 1/2</h3><h3></h3><h3> </h3><h3> at ^{2}$

$1400=15×40+ </p><p>1/2</p><p> </p><p> ×a×(40) </p><p> ^{2}$

$or, \\ 800=a×800 \\ </h3><h3></h3><h3>a=1m/s </h3><h3> ^{2}$

$hope \: its \: help \: u$

Answer 2

Answer:

I think you have got your answer