A truck left 30 minutes early than the scheduled time but in order to reach its destination 270 km away in time, it had to slow its usual speed by 6 km/hr. What is the usual speed of the truck?

Question

Prepp Team · Accepted Answer

Speed Calculation for Truck Journey

This problem involves calculating the usual speed of a truck based on information about its departure time, arrival time, distance, and a change in speed. We need to use the relationship between speed, distance, and time.

Defining Variables

Let's define the variables we'll use:

Let $d$ represent the distance of the journey, which is given as $d = 270$ km.
Let $S$ represent the usual speed of the truck in km/hr.
Let $T$ represent the usual time taken for the journey in hours.

From the basic formula, distance = speed × time, we have:

$d = S \times T$ $270 = S \times T$

We can express the usual time $T$ in terms of the usual speed $S$:

$T = \frac{270}{S}$

Understanding Time Adjustments

The problem states the truck left 30 minutes early and arrived exactly on time. Let's analyze the time implications:

The truck left 30 minutes ($\frac{1}{2}$ hour) earlier than the scheduled departure time.
The truck arrived at the originally scheduled arrival time.
This means the actual duration of the truck's journey was 30 minutes longer than it would have been if it had left on time and traveled at its usual speed.

So, the actual time taken ($T_{actual}$) is:

$T_{actual} = T + \frac{1}{2}$ hours.

Setting Up the Speed Equations

The problem also states that the truck had to slow its usual speed by 6 km/hr.

The actual speed ($S_{actual}$) was $S - 6$ km/hr.
The distance remains the same, $d = 270$ km.

Using the distance formula for the actual journey:

$d = S_{actual} \times T_{actual}$ $270 = (S - 6) \times (T + \frac{1}{2})$

Solving for the Usual Speed

Now, we substitute the expression for $T$ from the usual journey ($T = \frac{270}{S}$) into the equation for the actual journey:

$270 = (S - 6) \times (\frac{270}{S} + \frac{1}{2})$

Let's expand the equation:

$270 = S(\frac{270}{S}) + S(\frac{1}{2}) - 6(\frac{270}{S}) - 6(\frac{1}{2})$ $270 = 270 + \frac{S}{2} - \frac{1620}{S} - 3$

Subtract 270 from both sides:

$0 = \frac{S}{2} - \frac{1620}{S} - 3$

To eliminate the fractions and the negative terms, multiply the entire equation by $2S$:

$0 \times 2S = (\frac{S}{2} \times 2S) - (\frac{1620}{S} \times 2S) - (3 \times 2S)$ $0 = S^2 - 3240 - 6S$

Rearrange this into the standard quadratic equation form ($aS^2 + bS + c = 0$):

$S^2 - 6S - 3240 = 0$

We can solve this quadratic equation using the quadratic formula:

$S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a=1$, $b=-6$, and $c=-3240$. Plugging these values in:

$S = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-3240)}}{2(1)}$ $S = \frac{6 \pm \sqrt{36 + 12960}}{2}$ $S = \frac{6 \pm \sqrt{12996}}{2}$

Calculate the square root:

$\sqrt{12996} = 114$

Now substitute this back into the formula for $S$:

$S = \frac{6 \pm 114}{2}$

This gives two possible solutions for $S$:

$S_1 = \frac{6 + 114}{2} = \frac{120}{2} = 60$
$S_2 = \frac{6 - 114}{2} = \frac{-108}{2} = -54$

Since speed cannot be negative, we discard the negative solution ($S_2$). Therefore, the usual speed of the truck is $S = 60$ km/hr.

Verification of the Speed Result

Let's check if the calculated usual speed of 60 km/hr satisfies the conditions of the problem:

Usual Journey:
- Usual Speed ($S$) = 60 km/hr
- Distance ($d$) = 270 km
- Usual Time ($T$) = $\frac{d}{S} = \frac{270}{60} = 4.5$ hours.
Actual Journey:
- Actual Speed ($S_{actual}$) = $S - 6 = 60 - 6 = 54$ km/hr.
- Distance ($d$) = 270 km
- Actual Time ($T_{actual}$) = $\frac{d}{S_{actual}} = \frac{270}{54} = 5$ hours.

Now, let's compare the times. The actual journey took 5 hours, and the usual journey would have taken 4.5 hours. The difference is $5 - 4.5 = 0.5$ hours, which is exactly 30 minutes.

This confirms that if the truck traveled at 54 km/hr, the journey took 5 hours. If it had traveled at its usual speed of 60 km/hr, it would have taken 4.5 hours. Since it left 30 minutes early and arrived exactly on time, the total time elapsed from the scheduled departure was $T$ (the planned duration), but the actual travel time was $T + 0.5$ hours. The calculation holds true.

The usual speed of the truck is 60 km/hr.

Prepp Team · Answer

50 km/hr

Prepp Team · Answer

64 km/hr

Prepp Team · Answer

56 km/hr