Question 8e21a

Dec 7, 2016

$43.10 m$, rounded to two decimal places.

Explanation:

Constant speed of Car A $= 100 k m / h = 100 \times \frac{1000}{3600} = 27. \overline{7} m {s}^{-} 1$
Let us consider Car B first.
It enters the acceleration lane at a speed $= 25 k m / h r = 6.9 \overline{4} m {s}^{-} 1$.

It accelerates uniformly and enters the main traffic lane after traveling $70 m \text{ in } 5 s$.
Using the following kinematic equation to obtain acceleration $a$
$s = u t + \frac{1}{2} a {t}^{2}$
$70 = 6.9 \overline{4} \times 5 + \frac{1}{2} a \times {5}^{2}$
$\implies \frac{1}{2} a \times {5}^{2} = 70 - 6.9 \overline{4} \times 5$
Dividing both sides by $5$ we get
$\frac{5}{2} a = 14 - 6.9 \overline{4}$
$\implies a = 2.8 \overline{2} m {s}^{-} 2$

To calculate the final speed at the time of entering the main traffic we use the following kinematic equation
$v = u + a t$
$v = 6.9 \overline{4} + 2.8 \overline{2} \times 5$
$\implies v = 21.0 \overline{5} m {s}^{-} 1$

It then continues to accelerate at the same rate until it reaches a speed of $27. \overline{7} m {s}^{-} 1$, which it then maintains.

Time taken to reach top speed is calculated from the kinematic equation
$v = u + a t$
$27. \overline{7} = 6.9 \overline{4} \times 2.8 \overline{2} t$

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

$= \frac{v + u}{2} = \frac{6.9 \overline{4} + 27. \overline{7}}{2} = 17.36 \overline{1} m {s}^{-} 1$

Distance traveled in the main traffic lane$= 17.36 \overline{1} \times 7.382 = 128.16 m$, rounded to 2 decimal places

Now Car A

Distance traveled during this time period $= 27. \overline{7} \times 7.382 = 205.0 \overline{5} m$

As the car A was $120 m$ 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