for car A
initial velocity u=0m/sec
acceleration=a=2m/s2
speed attained by A =27 m/sec
time taken by A to attain the speed =t
t=[v-u]/a=[27-0]/2=13.5m/sec2
distance traveled by car A in this time =s=ut+1/2at2
s= 0+0.5*2*13.5^2=182.25m
in this time distance traveled by B towards A=27*20/2=270m
so distance remaining between A and B after 27/2sec=2000-[182.25+270]=1547.75m
relative speed between A and B=27-[-20]=47m/sec
as they are moving in opposite direction.
now time taken by A to cross B =1547.75/47=33sec
in this time distance traveled by A=27*33=891m
hence the distances traveled by car A when they pass each
other=182.25+891=1073.25 m
Car A starts from rest at t = 0 and travels along a straight road with a constant
acceleration of 2 m/s2 until it reaches a speed of 27 m/s. Afterwards it maintains this
speed. Also, when t = 0, car B located 2000 m down the road is traveling towards A at a
constant speed of 20 m/s. Determine the distance traveled by car A when they pass each
other.
3 answers
Two cars are traveling along a straight road. Car a maintains a constant speed of 89 km/hr and car b maintains a constant speed of 119 km/hr. at t=0 car b is behind 38km from car a. How far will car a travel from t =0 before it is overtaken by car b. how long from t=0 will it take for this to happen
morty let's go on an adventure
3 answers