A commuter train travels between two downtown stations....

A commuter train travels between two downtown stations. Because the stations are only $1.00 \mathrm{km}$ apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the time interval $\Delta t$ between two stations by accelerating for a time interval $\Delta t_{1}$ at a rate $a_{1}=0.100 \mathrm{m} / \mathrm{s}^{2}$ and then immediately braking with acceleration $a_{2}=-0.500 \mathrm{m} / \mathrm{s}^{2}$ for a time interval $\Delta t_{2} .$ Find the minimum time interval of travel $\Delta t$ and the time interval $\Delta t_{1}$.

Key Concepts

Optimization under Kinematic Constraints

In scenarios where a minimum travel time is sought, such as this problem, it is necessary to optimize the parameters (e.g., the durations of acceleration and deceleration) subject to the constraints of the given distance and the constant accelerations. This requires using the appropriate kinematic relationships and equations to derive conditions that minimize the total travel time while ensuring that the distance covered in both motion segments adds up to the total trip length.

Velocity Matching at Phase Transition

At the point where one motion phase ends and another begins, the speed at the end of the acceleration phase must equal the speed at the beginning of the deceleration phase. This continuity condition provides an essential relationship between the time spent in each phase (or the distances covered) and is used to link the acceleration and deceleration segments together within the overall journey.

Constant Acceleration Kinematics

This concept involves the kinematic equations that relate displacement, velocity, acceleration, and time under conditions of uniform acceleration. In problems where an object accelerates (or decelerates) at a constant rate, these equations allow us to calculate how far the object travels during the acceleration period, what its final speed is, and how long it takes to reach a certain velocity or cover a specific distance.

Segmented or Piecewise Motion

This idea refers to breaking down motion into distinct segments, each with its own constant acceleration. In the context of the problem, the train’s journey is divided into an acceleration phase and a deceleration phase, with each phase having its own acceleration value and time interval. Analyzing each segment separately while ensuring the continuity of motion (such as matching speeds at phase transitions) is crucial for solving segmented motion problems.

Transcript

00:01 It initially accelerates with an acceleration of 0 .1 meters per square second over time t1 and then decelerates with a deceleration of minus 0 .5 meters per square second over time t2.

00:14 So we can use the equation that x is equal to a half a1, t1 squared, since we assume it starts from rest, plus its initial velocity for the second part of the journey, which is b1, v1 times time of t2 which is the time of way to decelerate plus a half a 2 t2 t2 square so we combine the distance covered for the first part of the journey and the second part of the journey in one equation we also know that the total time t for the journey is equal to t1 plus time t2 and that v1 is simply a1 times t1 and this is the initial speed for the second part of the journey and since the train comes to rest that's equal to minus a 2 times t 2 so we can substitute these expressions in our equation for x and so x we know to be 1 ,000 meters so we substitute us in so 1 ,000 is equal to a half a 1 t1 square plus v1 which we see from above is a1 times t1 times t 2 which is minus a1 times t 1 over a 2 plus a half a 2 and t2 simply a 1 t1 divided by a 2 again all squared and that's minus so simplifying this we get that 1000, which x is equal to a half a 1 into 1 minus a 1 over 8 2 all times t1 squared.

03:05 And so this makes our expression a lot simpler.

03:10 And so now we can substitute values since we know our accelerations and we're left from one unknown, namely t1.

03:17 So 1000 is equal to a half times a 1, which we know is 0 .1 meters per square second into 1 minus 0 .1 over minus 0 .5 all times t1 squared.

03:48 And so if you rearrange the equation we can solve for time t1 and t1 can be seen to be square root of 20 ,000 divided by 1 .2 and we get the time t1 for the acceleration to be 129 seconds...

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