two buses A and B leave the same station bus A heads due east while bus B heads due north the north bound bus travelled 10km\h faster than the east bound bus after 5hours the two buses are 250km apart calculate the speed of the north bound bus

To solve this problem, let's denote:

- The speed of bus A (eastbound) as \( v_A \) km/h.
- The speed of bus B (northbound) as \( v_B = v_A + 10 \) km/h.

After 5 hours, the distance travelled by bus A is:
\[ d_A = v_A \times 5 \]

And the distance travelled by bus B is:
\[ d_B = v_B \times 5 = (v_A + 10) \times 5 \]

Given that after 5 hours, the distance between the two buses is 250 km, we can use the Pythagorean theorem to relate these distances. The situation forms a right triangle with the legs representing the distances travelled by each bus and the hypotenuse representing the distance between the buses.

So, we have:
\[ d_A^2 + d_B^2 = 250^2 \]

Substitute the distances:
\[ (5v_A)^2 + [5(v_A + 10)]^2 = 250^2 \]

Simplify the equation:
\[ 25v_A^2 + 25(v_A + 10)^2 = 62500 \]

Expanding the squared term:
\[ 25v_A^2 + 25(v_A^2 + 20v_A + 100) = 62500 \]

Combine like terms:
\[ 25v_A^2 + 25v_A^2 + 500v_A + 2500 = 62500 \]

Simplify:
\[ 50v_A^2 + 500v_A + 2500 = 62500 \]

Divide the entire equation by 50:
\[ v_A^2 + 10v_A + 50 = 1250 \]

Subtract 1250 from both sides:
\[ v_A^2 + 10v_A - 1200 = 0 \]

This is a quadratic equation in the form \( ax^2 + bx + c = 0 \). To solve it, we can use the quadratic formula:
\[ v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where \( a = 1 \), \( b = 10 \), and \( c = -1200 \):
\[ v_A = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-1200)}}{2 \cdot 1} \]
\[ v_A = \frac{-10 \pm \sqrt{100 + 4800}}{2} \]
\[ v_A = \frac{-10 \pm \sqrt{4900}}{2} \]
\[ v_A = \frac{-10 \pm 70}{2} \]

This gives us two solutions:
\[ v_A = \frac{60}{2} = 30 \]
\[ v_A = \frac{-80}{2} = -40 \]

Since a negative speed doesn't make sense in this context, we discard \( v_A = -40 \). Thus, \( v_A = 30 \) km/h.

The speed of the northbound bus B is:
\[ v_B = v_A + 10 = 30 + 10 = 40 \text{ km/h} \]

Therefore, the speed of the northbound bus is \( 40 \) km/h.

Would you like details or have any questions?

Here are 8 related questions to expand on this information:

1. How far did bus A travel in 5 hours?
2. How far did bus B travel in 5 hours?
3. What would be the speed of the eastbound bus if the northbound bus travelled 50 km/h?
4. How would the problem change if bus A travelled north and bus B travelled east?
5. What if the buses travelled for 10 hours instead of 5?
6. How does the Pythagorean theorem apply to this problem?
7. What if the speed difference between the two buses was 20 km/h instead of 10 km/h?
8. What are other methods to solve such distance and speed problems?

**Tip:** When solving word problems, clearly define all variables and systematically apply relevant equations and theorems to find the solution.

Learn how to calculate the speed of the northbound bus B when given the speed of eastbound bus A, using distance, speed, and time concepts. This problem involves applying the Pythagorean theorem and solving a quadratic equation. Suitable for students in Grades 9-12.

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