Question

Starting at station \(\mathrm{A}\), a commuter train accelerates at 3 meters per second per second for 8 seconds, then travels at constant speed \(v_{m}\) for 100 seconds, and finally brakes (decelerates) to a stop at station \(\mathrm{B}\) at 4 meters per second per second. Find (a) \(v_{m}\) and (b) the distance between \(\mathrm{A}\) and \(\mathrm{B}\).

Short answer

(a) 24 m/s; (b) 2568 meters.

Step-by-step solution

Calculate velocity after acceleration

The train accelerates from rest, so the initial velocity is 0. Use the formula \( v = u + at \) to find the velocity after 8 seconds, where \( u = 0 \), \( a = 3 \text{ m/s}^2 \), and \( t = 8 \text{ s} \). \[ v = 0 + (3 \times 8) = 24 \text{ m/s} \]Thus, the maximum velocity \( v_m \) is 24 m/s.

Calculate distance during acceleration

Use the formula \( s = ut + \frac{1}{2} a t^2 \) to calculate the distance covered during the acceleration phase. Here \( u = 0 \), \( a = 3 \text{ m/s}^2 \), and \( t = 8 \text{ s} \).\[ s = 0 + \frac{1}{2} \times 3 \times 8^2 = 96 \text{ m} \]

Calculate distance at constant speed

Once the train reaches velocity \( v_m = 24 \text{ m/s} \), it travels at this constant speed for 100 seconds. The distance is calculated using \( s = vt \).\[ s = 24 \times 100 = 2400 \text{ m} \]

Calculate distance during deceleration

The train decelerates at \( a = -4 \text{ m/s}^2 \) from \( v = 24 \text{ m/s} \) to \( v = 0 \text{ m/s} \). Use \( v^2 = u^2 + 2as \) to find the stopping distance. Here \( v = 0 \), \( u = 24 \text{ m/s} \), and \( a = -4 \text{ m/s}^2 \).\[ 0 = 24^2 + 2(-4)s \]Solving for \( s \):\[ 576 = 8s \]\[ s = 72 \text{ m} \]

Calculate total distance between stations

Add the distances from each phase to find the total distance between Stations A and B. \[ \text{Total distance} = 96 + 2400 + 72 = 2568 \text{ m} \]

Key concepts

Acceleration

Acceleration is the rate of change of velocity of an object with time. To calculate acceleration, you need to know the change in velocity and the time over which this change happens.

• The formula to calculate acceleration is: \[ a = \frac{{\Delta v}}{{\Delta t}} \]Here, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the time period over which this change occurs.
• Acceleration can be positive, which means an increase in speed, or negative (also known as deceleration), which indicates a decrease in speed.
• In the exercise, the train starts from rest (initial velocity \( u = 0 \)) and accelerates at \( 3 \text{ m/s}^2 \) for 8 seconds. This results in a change in velocity (\( \Delta v \)) to \( 24 \text{ m/s} \).
• It's important to understand that acceleration doesn't just apply to increases in speed; it's any change in velocity, including slowing down or changing direction.

Velocity

Velocity is a vector quantity that refers to the rate at which an object changes its position. It includes both the speed and direction of the object's movement.

• The formula for calculating velocity after a period of constant acceleration is: \[ v = u + at \]Where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
• In the problem, once the train reaches its maximum speed after acceleration, this constant speed \( v_m \) is \( 24 \text{ m/s} \), which it maintains for a certain period.
• Velocity is crucial in determining how long it will take to travel a certain distance with constant speed.

Besides speed, remember that velocity takes the direction into account. That means if two objects are moving at the same speed but in opposite directions, they have different velocities.

Distance Calculation

Calculating distance is an essential part of kinematics problems. Distance refers to the total movement of an object, regardless of its direction.

• To calculate the distance during acceleration, the formula is: \[ s = ut + \frac{1}{2} a t^2 \]Where \( s \) is the distance, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
• For a period of constant speed, the distance is calculated using: \[ s = vt \]Here, \( v \) is the velocity and \( t \) is the time.
• These formulas help you calculate how far an object traveled during various phases of motion. In the example, the train travels an extra 2400 meters at a constant speed after accelerating.
• Always consider each phase of motion separately to get the most accurate distance calculations.

Deceleration

Deceleration is the process of slowing down or decreasing speed. It is essentially negative acceleration, and like acceleration, it involves a change in velocity over time.

• The formula to calculate deceleration when stopping is the same as for acceleration: \[ v^2 = u^2 + 2as \]Where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the deceleration, and \( s \) is the stopping distance.
• In the current exercise, the train decelerates at \( -4 \text{ m/s}^2 \) until coming to a rest, which requires calculating the stopping distance.
• Note that deceleration can never cause an object to go faster in its initial direction, unlike acceleration.
• Understanding deceleration is crucial for accurately predicting how long it takes for an object to stop moving once it begins slowing down.