Question : On a two lane car road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C. What minimum acceleration of Car B is required to avoid an accident?
Doubt by Suhani
Solution :
Here
uB=54 km/h
uB=54×(5/18)
uB=15 m/s
uB=15 m/s
uA=36 km/h
uA=36×(5/18)
uA=10 m/s
uA=36×(5/18)
uA=10 m/s
Similarly
uC=-54 km/h
uC=-15 m/s
uC=-54 km/h
uC=-15 m/s
Distance = AB=AC=1 km = 1000 m
Relative Velocity of Car B w.r.t. A
uBA=uB-uA
uBA=uB-uA
uBA=15-(10)
uBA=5 m/s
uBA=5 m/s
Relative Velocity of Car C w.r.t. A
uCA=uC-uA
uCA=uC-uA
uCA=-15-(10)
uCA=-25 m/s
uCA=-25 m/s
Time taken by car C to overtake Car A
s=ut+½at²
s=ut+½at²
AC=uCA(t)+½[aCA](t)
1000=(|-25|)t+½[0]t
1000/25=t
t= 40 sec
1000=(|-25|)t+½[0]t
1000/25=t
t= 40 sec
Hence the time taken by car B should be 40 sec or less if it want to overtake car A before car C.
For Car B
s=ut+½at²
s=ut+½at²
AB=uBAt+½a(t)²
1000=5(40)+½a(40)²
1000=200+½a(1600)
1000=5(40)+½a(40)²
1000=200+½a(1600)
1000-200=800a
800=800a
800/800=a
1=a
a=1 m/s²
800=800a
800/800=a
1=a
a=1 m/s²
Hence, the required minimum acceleration of car B should be 1 m/s².