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: Set up the general form of a linear function.
Since the speed changes at a steady (linear) rate, we can write it as \(v(t) = v_0 + at\), where \(v_0\) is the speed at \(t=0\) and \(a\) is the constant rate of change per minute.

Step 2: Find \(v_0\) from the table.
At \(t=0\), the table shows speed equal to 40 km/hour, so \(v_0 = 40\).

Step 3: Find \(a\) using a second data point.
At \(t=60\), the table gives speed = 50 km/hour. Substitute into \(v(t) = 40+at\):
\[ 40 + 60a = 50 \implies 60a = 10 \implies a = \frac{1}{6} \]
So the formula becomes \(v(t) = 40+\frac{t}{6}\).

Step 4: Check the formula against other rows of the table.
At \(t=90\): \(40+90/6=40+15=55\), matches the table. At \(t=150\): \(40+150/6=40+25=65\), matches the table. At \(t=45\): \(40+45/6=40+7.5=47.5\), matches the table too. The formula fits every entry.

Step 5: Rule out the wrong options.
Option \(t/6\) gives 0 at \(t=0\), not 40, so it is wrong. Option \(6t\) also gives 0 at \(t=0\). Option \(40+t\) would give \(40+30=70\) at \(t=30\), but the table shows 45 at \(t=30\), so it is wrong too.

Step 6: Final Answer.
The formula that fits every point in the table is \(40+\frac{t}{6}\). \[ \boxed{40+\frac{t}{6}} \]