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A train travelling at a uniform speed for 360km would have taken 48 minutes less to travel the same distance if its speed were 5km// hour more. Find the original speed of the train.

A train travelling at a uniform speed for 360 km 360 \mathrm{~km} would have taken 4848 minutes less to travel the same distance if its speed were 5 km/ 5 \mathrm{~km} / hour more. Find the original speed of the train.

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Relate position, velocity, speed, and acceleration using derivativesright chevron icon

Full solution

Q. A train travelling at a uniform speed for 360 km 360 \mathrm{~km} would have taken 4848 minutes less to travel the same distance if its speed were 5 km/ 5 \mathrm{~km} / hour more. Find the original speed of the train.
  1. Denote original speed: Let's denote the original speed of the train as 'ss' km/hour. The time taken to travel 360360 km at this speed is 360s\frac{360}{s} hours. If the speed were increased by 55 km/hour, the new speed would be 's+5s + 5' km/hour, and the time taken to travel the same distance would be 360s+5\frac{360}{s + 5} hours. According to the problem, the time difference is 4848 minutes, which is 4860\frac{48}{60} hours. We can set up the equation:\newline360s−360s+5=4860\frac{360}{s} - \frac{360}{s + 5} = \frac{48}{60}
  2. Calculate time difference: First, we need to simplify the equation. To do this, we find a common denominator for the fractions on the left side of the equation, which is s(s+5)s(s + 5). This gives us:\newline(360(s+5)−360s)/s(s+5)=48/60(360(s + 5) - 360s) / s(s + 5) = 48/60
  3. Simplify the equation: Now, we simplify the numerator on the left side of the equation:\newline360s+1800−360ss(s+5)=4860360s + 1800 - \frac{360s}{s(s + 5)} = \frac{48}{60}\newlineThe '360s360s' terms cancel out, leaving us with:\newline1800s(s+5)=4860\frac{1800}{s(s + 5)} = \frac{48}{60}
  4. Simplify numerator: Next, we simplify the right side of the equation by reducing the fraction 4860\frac{48}{60}:\newline4860=45\frac{48}{60} = \frac{4}{5}\newlineSo, our equation now is:\newline1800s(s+5)=45\frac{1800}{s(s + 5)} = \frac{4}{5}
  5. Reduce fraction: To solve for ss, we cross-multiply: 1800×5=4×s(s+5)1800 \times 5 = 4 \times s(s + 5) 9000=4s2+20s9000 = 4s^2 + 20s
  6. Cross-multiply: We now have a quadratic equation. To solve for ss, we need to set the equation to zero: 4s2+20s−9000=04s^2 + 20s - 9000 = 0
  7. Set to zero: We can simplify the equation by dividing all terms by 44: s2+5s−2250=0s^2 + 5s - 2250 = 0
  8. Factor quadratic equation: Now, we need to factor the quadratic equation or use the quadratic formula to find the value of 's'. The quadratic formula is:\newlines=−b±b2−4ac2as = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\newlineIn our equation, a=1a = 1, b=5b = 5, and c=−2250c = -2250. Let's calculate the discriminant b2−4acb^2 - 4ac first:\newlineDiscriminant = 52−4(1)(−2250)=25+9000=90255^2 - 4(1)(-2250) = 25 + 9000 = 9025
  9. Calculate discriminant: Since the discriminant is positive, we have two real solutions. Now we can find the values of ss:s=−5±90252s = \frac{{-5 \pm \sqrt{9025}}}{{2}}
  10. Find values of \newliness: Calculating the square root of \newline90259025 gives us \newline9595. So, the equation becomes:\newline\newlines=(−5±95)/2s = (-5 \pm 95) / 2
  11. Calculate square root: We have two possible solutions for ss:s=−5+952s = \frac{{-5 + 95}}{2} or s=−5−952s = \frac{{-5 - 95}}{2}s=902s = \frac{90}{2} or s=−1002s = \frac{-100}{2}s=45s = 45 or s=−50s = -50
  12. Two possible solutions: Since speed cannot be negative, we discard the negative solution. Therefore, the original speed of the train is 45 km/hour45\,\text{km/hour}.

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