Car a leaves city c at 5 pm and drives at a speed of 40 kmph. 2 hours later another car b leaves city c and drives in the same direction as car
a. in how much time will car b be 9 km ahead of car a if speed of car b is 60 kmph?
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car A started 2 hours before ,so distance covered by car A when Car B starts is:
40*2=80 km(dis= speed*time)
now make a equation using distance(equating both distance ):
car A cover 80 km and going in 40 kmph, car B is starting with 60 kmph-
80+y*40=y*60, y is the time taken by the cars(distance covered by cars are equal)
please, understand that both cars are travelling ,but car B have to go 9 km apart, so equating time relation of both cars will help:
w/40=(w+9)/60
w=18 km (after travelling 18 km by car A , car b will be 9 km ahead)
now cal the time
speed*time=distance
40*t=18
t=.45*60(convert into mins)
t=27
so the answer is 4 hr,27mins
solutions also can be derived using relative speed
40*2=80 km(dis= speed*time)
now make a equation using distance(equating both distance ):
car A cover 80 km and going in 40 kmph, car B is starting with 60 kmph-
80+y*40=y*60, y is the time taken by the cars(distance covered by cars are equal)
please, understand that both cars are travelling ,but car B have to go 9 km apart, so equating time relation of both cars will help:
w/40=(w+9)/60
w=18 km (after travelling 18 km by car A , car b will be 9 km ahead)
now cal the time
speed*time=distance
40*t=18
t=.45*60(convert into mins)
t=27
so the answer is 4 hr,27mins
solutions also can be derived using relative speed
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