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). At a certain instant, when the distance AB is equal to AC, both 1 km B decided to overtake A before Cdoes. What minimum acceleration of car B is required to avoid and accident?

Answer 1

Answer:

Speed of A=36km/hr=36×  

18

5

​  

m/s=10m/s

Speed of B= Speed of C=54×  

18

5

​  

m/s=15m/s

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

Time taken by C to overtake A=  

25

1000

​  

 

=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+  

2

1

​  

at  

2

 

1400=15×40+  

2

1

​  

×a×(40)  

2

 

or, 800=a×800

a=1m/s  

2

  Ans

Explanation: