Question #8e21a

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