Car $\mathrm{X}$ and Car $\mathrm{Y}$ are travelling in the same direction in two different lanes on a long straight highway. Car $\mathrm{X}$ is travelling at a constant speed of $90 \mathrm{~km} / \mathrm{h}$ and has a length of $5 \mathrm{~m}$. Car $\mathrm{Y}$ is travelling at a constant speed of $91 \mathrm{~km} / \mathrm{h}$ and has a length of $6 \mathrm{~m}$. Car Y starts behind Car X and eventually passes Car X. The length of time between the instant when the front of Car Y is lined up with the back of Car X and the instant when the back of Car $\mathrm{Y}$ is lined up with the front of Car $\mathrm{X}$ is $t$ seconds. The value of $t$ is
(A) 39.6
(B) 18.0
(C) 21.6
(D) 46.8
(E) 32.4

## Part C: Each correct answer is worth 8.
Since there are 60 seconds in 1 minute, then $t$ seconds is equivalent to $\frac{t}{60}$ minutes.

Since there are 60 minutes in 1 hour, then $\frac{t}{60}$ minutes is equivalent to $\frac{t}{60 \times 60}$ hours or $\frac{t}{3600}$ hours.

Consider the distances that Car X and Car Y travel between the instant when the front of Car Y is lined up with the back of Car X and the instant when the back of Car Y is lined up with the front of Car X.

Since the length of Car X is $5 \mathrm{~m}$ and the length of Car Y is $6 \mathrm{~m}$, then during this interval of time, Car Y travels $5+6=11 \mathrm{~m}$ farther than Car X. (The front of Car Y must, in some sense, travel all of the way along the length of the Car X and be $6 \mathrm{~m}$ ahead of the front of Car $\mathrm{X}$ so that the back of Car Y is lined up with the front of Car X.)

Since there are 1000 metres in $1 \mathrm{~km}$, then $11 \mathrm{~m}$ is equivalent to $0.011 \mathrm{~km}$.

Since Car X travels at $90 \mathrm{~km} / \mathrm{h}$, then in $\frac{t}{3600}$ hours, Car X travels $\frac{90 t}{3600} \mathrm{~km}$.

Since Car Y travels at $91 \mathrm{~km} / \mathrm{h}$, then in $\frac{t}{3600}$ hours, Car Y travels $\frac{91 t}{3600} \mathrm{~km}$.

Therefore, $\frac{91 t}{3600}-\frac{90 t}{3600}=0.011$, or $\frac{t}{3600}=0.011$ and so $t=3600 \times 0.011=36 \times 1.1=39.6$.