A train initially travels at 30 m/s. After applying the brakes, its velocity decreases by 10% in the first 40 seconds. We need to: i. Calculate the velocity at the end of a further 80 seconds, assuming the retardation is proportional to the velocity during the whole 120 second period. ii. Derive an expression for the retarding force per ton mass of the train in terms of its velocity. iii. Find the power being dissipated at the end of the 120 second period if the train has a mass of 500 tons.

1. Problem Description

A train initially travels at 30 m/s. After applying the brakes, its velocity decreases by 10% in the first 40 seconds. We need to:

i. Calculate the velocity at the end of a further 80 seconds, assuming the retardation is proportional to the velocity during the whole 120 second period.

ii. Derive an expression for the retarding force per ton mass of the train in terms of its velocity.

iii. Find the power being dissipated at the end of the 120 second period if the train has a mass of 500 tons.

2. Solution Steps

i. Calculate the velocity at the end of 120 s.

First, we find the velocity after 40 seconds.

$v_1 = 30 - 0.10 \times 30 = 30 - 3 = 27$ m/s.

Since the retardation (deceleration) is proportional to the velocity, we have:

$a = -kv$

where $a$ is acceleration, $v$ is velocity, and $k$ is a constant of proportionality.

Since $a = \frac{dv}{dt}$, we have:

$\frac{dv}{dt} = -kv$

Separating variables and integrating:

$\int \frac{dv}{v} = \int -k dt$

$\ln(v) = -kt + C$

$v = e^{-kt + C} = e^C e^{-kt} = A e^{-kt}$

At $t = 0$, $v = 30$ m/s, so $A = 30$. Thus,

$v = 30 e^{-kt}$

At $t = 40$ s, $v = 27$ m/s. Substituting these values into the equation:

$27 = 30 e^{-40k}$

$e^{-40k} = \frac{27}{30} = \frac{9}{10} = 0.9$

$-40k = \ln(0.9)$

$k = -\frac{\ln(0.9)}{40} \approx 0.00263$

We want to find the velocity at $t = 120$ s:

$v = 30 e^{-120k} = 30 e^{-120(-\frac{\ln(0.9)}{40})} = 30 e^{3\ln(0.9)} = 30 (0.9)^3 = 30 \times 0.729 = 21.87$ m/s.

ii. Derive an expression for the retarding force in N/ton mass of the train.

The retarding force $F = ma$, where $m$ is the mass and $a$ is the acceleration. We know that $a = -kv$.

Given that the mass is in tons, we have the force per ton as:

$F/m = a = -kv = \frac{\ln(0.9)}{40} v$

The retarding force per ton is:

$F_{ton} = -kv = \frac{\ln(0.9)}{40} v \approx -0.00263v$ N/ton. Since we are asked for retarding force, it should be positive.

$F_{ton} = -\frac{\ln(0.9)}{40} v$ N/ton

iii. Find the power being dissipated at the end of the whole period.

The power dissipated is given by $P = Fv$, where $F$ is the retarding force.

The mass of the train is 500 tons. The velocity at $t=120$ s is $21.87$ m/s.

First, we find the total retarding force:

$F = 500 \times 1000 \times a = 500 \times 1000 \times (-kv) = 500000 \times (\frac{\ln(0.9)}{40}) \times 21.87$

$F = 500000 \times 0.00263 \times 21.87 = 28757.55$ N

$P = Fv = 28757.55 \times 21.87 = 628994.99 \approx 628995$ W.

3. Final Answer

i. The velocity at the end of 120 s is 21.87 m/s.

ii. The retarding force per ton is $-\frac{\ln(0.9)}{40} v$ N/ton.

iii. The power being dissipated at the end of the whole period is approximately 628995 W.