Question #8e21a

1 Answer
Dec 7, 2016

Answer:

#43.10 m#, rounded to two decimal places.

Explanation:

Constant speed of Car A #=100km//h=100xx1000/3600=27.bar7ms^-1#
Let us consider Car B first.
It enters the acceleration lane at a speed #=25 km//hr = 6.9bar4 ms^-1#.

It accelerates uniformly and enters the main traffic lane after traveling #70 m" in "5s#.
Using the following kinematic equation to obtain acceleration #a#
#s=ut+1/2at^2#
#70=6.9bar4xx5+1/2axx5^2#
#=>1/2axx5^2=70-6.9bar4xx5#
Dividing both sides by #5# we get
#5/2a=14-6.9bar4#
#=>a=2.8bar2ms^-2#

To calculate the final speed at the time of entering the main traffic we use the following kinematic equation
#v=u+at#
#v=6.9bar4+2.8bar2xx5#
#=>v=21.0bar5ms^-1#

It then continues to accelerate at the same rate until it reaches a speed of #27.bar7ms^-1#, which it then maintains.

Time taken to reach top speed is calculated from the kinematic equation
#v=u+at#
#27.bar7 = 6.9bar4 xx 2.8bar2 t#

#=>t=(27.bar7 – 6.9bar4) / {2.8bar2 } = 7.382 s#, rounded to three decimal places.
During this period of acceleration average speed

#=(v+u)/2=(6.9bar4 + 27.bar 7 ) / 2 = 17.36bar1 ms^-1#

Distance traveled in the main traffic lane#=17.36bar1 xx 7.382 = 128.16 m#, rounded to 2 decimal places

Now Car A

Distance traveled during this time period #=27.bar7 xx 7.382 = 205.0bar5m#

As the car A was #120m# behinf car B when car B entered the main traffic lane. Therefore, distance of car A from car B
#=120 + 128.16– 205.0bar5 m = 43.10 m#, rounded to two decimal places