Question

Use the same train-speed table as above.

Time (minutes)	0	30	45	60	90	120	150	180
Speed (km/hour)	40	45	47.5	50	55	60	65	70

At time \(t\) (minutes) after the beginning, which formula fits the train's speed according to the table (assume the change is linear over time)?

• A. \(\dfrac{t}{6}\)
• B. \(6t\)
• C. \(40+t\)
• D. \(40+\dfrac{t}{6}\)

Hint:

Use the speed at \(t=0\) as the starting value, then find the constant rate of change per minute from any other row in the table.

Answer 1

The Correct Option is D

Step 1: Work out the rate of change per minute first.
Over the full 3-hour (180-minute) span, speed goes from 40 km/hour to 70 km/hour, a rise of 30 km/hour in 180 minutes. That is a rate of $30/180 = 1/6$ km/hour for every minute that passes.

Step 2: Write the speed as starting speed plus this rate times time.
\[ v(t) = 40 + \frac{1}{6}t \]
This says the speed starts at 40 km/hour when $t=0$ and climbs by $1/6$ km/hour for every minute.

Step 3: Test it on a value not used to build the formula.
Try $t=120$: $40+120/6=40+20=60$, which matches the table's entry of 60 km/hour at 120 minutes. Try $t=45$: $40+45/6=47.5$, which also matches.

Step 4: Eliminate options that fail the $t=0$ check.
At $t=0$, the actual speed is 40 km/hour. Options $t/6$ and $6t$ both give 0 at $t=0$, so neither can be right. Option $40+t$ gives 70 km/hour already at $t=30$, far above the table's 45 km/hour, so it also fails.

Step 5: Final Answer.
Only $40+\frac{t}{6}$ survives every check.
\[ \boxed{40+\frac{t}{6}} \]