1. **State the problem:** A train starts from rest and accelerates uniformly for 5 minutes to reach a speed $v$ km/hr. It then travels at this constant speed for 45 minutes and finally decelerates uniformly to rest in 10 minutes. The total distance traveled in 1 hour is 49 km. We need to find the maximum speed $v$ and the average acceleration during the first 5 minutes. 2. **Relevant formulas and rules:** - Distance traveled is the area under the speed-time graph. - Average acceleration $a = \frac{\Delta v}{\Delta t}$. - Time must be converted to hours since speed is in km/hr. 3. **Calculate the total distance using the trapezoid area:** - Acceleration phase (0 to 5 min): speed increases from 0 to $v$ km/hr. - Constant speed phase (5 to 50 min): speed is $v$ km/hr. - Deceleration phase (50 to 60 min): speed decreases from $v$ to 0. Convert minutes to hours: $$5 \text{ min} = \frac{5}{60} = \frac{1}{12} \text{ hr}, \quad 45 \text{ min} = \frac{45}{60} = \frac{3}{4} \text{ hr}, \quad 10 \text{ min} = \frac{10}{60} = \frac{1}{6} \text{ hr}$$ 4. **Calculate distance for each phase:** - Acceleration phase distance (triangle): $$d_1 = \frac{1}{2} \times v \times \frac{1}{12} = \frac{v}{24}$$ - Constant speed phase distance (rectangle): $$d_2 = v \times \frac{3}{4} = \frac{3v}{4}$$ - Deceleration phase distance (triangle): $$d_3 = \frac{1}{2} \times v \times \frac{1}{6} = \frac{v}{12}$$ 5. **Sum distances and set equal to total distance:** $$d_1 + d_2 + d_3 = 49$$ $$\frac{v}{24} + \frac{3v}{4} + \frac{v}{12} = 49$$ 6. **Find common denominator and simplify:** Common denominator is 24: $$\frac{v}{24} + \frac{18v}{24} + \frac{2v}{24} = 49$$ $$\frac{v + 18v + 2v}{24} = 49$$ $$\frac{21v}{24} = 49$$ 7. **Solve for $v$:** $$21v = 49 \times 24$$ $$v = \frac{49 \times 24}{21}$$ Simplify numerator and denominator: $$v = \frac{49 \times \cancel{24}}{\cancel{21}} \times \frac{24}{21} = \frac{49 \times 24}{21}$$ Calculate: $$v = \frac{49 \times 24}{21} = 49 \times \frac{24}{21} = 49 \times \frac{8}{7} = 7 \times 8 = 56$$ So, the maximum speed $v$ is 56 km/hr. 8. **Calculate average acceleration during first 5 minutes:** $$a = \frac{\Delta v}{\Delta t} = \frac{56 - 0}{\frac{1}{12}} = 56 \times 12 = 672 \text{ km/hr}^2$$ **Final answers:** - Maximum speed attained: **56 km/hr** - Average acceleration during first 5 minutes: **672 km/hr²**