Q#66
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations:
(a) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and
(b) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate of 1.1 $m/s^2$ until it reaches 95 km/h then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at $-2 \ m/s^2$. Assume it stops at each intermediate station for 22 s.
Answer:
This problem can be analyzed as a series of three one-dimensional motions: the acceleration phase, the constant-speed phase, and the deceleration phase.
The maximum speed of the train is as follows:
$(95 \ km/h)\left(\frac{1000 \ m/km}{3600 \ s/h}\right)=26.39 \ m/s$
In the acceleration phase, the initial velocity is v = 0, the acceleration is a = 1.1 $m/s^2$, and the final velocity is v = 26.39 m/s.
Find the elapsed time for the acceleration phase from Eq. $v=v_0+at$.
$t_{acc}=\frac{v-v_0}{a}$
$t_{acc}=\frac{26.39 \ m/s-0}{1.1 \ m/s^2}=23.99 \ s$
Find the displacement during the acceleration phase from Eq.
$x-x_0=v_0t+\frac{1}{2}at^2$ (*)
$(x-x_0)_{acc}=v_0t_{acc}+\frac{1}{2}at_{acc}^2$
$(x-x_0)_{acc}=0+\frac{1}{2}(1.1 \ m/s^2)(23.99 \ s)^2=316.5 \ m$
In the deceleration phase, the initial velocity is $v_0$ = 26.39 m/s, the acceleration is a = −2.0 $m/s^2$, and the final velocity is v = 0.
Find the elapsed time for the deceleration phase from Eq. $v=v_0+at$.
$t_{dec}=\frac{v-v_0}{a}$
$t_{dec}=\frac{0-26.39 \ m/s}{-2.0 \ m/s^2}=13.20 \ s$
Find the distance traveled during the deceleration phase from Eq. (*)
$(x-x_0)_{dec}=v_0t_{dec}+\frac{1}{2}at_{dec}^2$
$(x-x_0)_{dec}=(26.39 \ m/s)(13.20 \ s)+\frac{1}{2}(-2.0 \ m/s^2)(13.20 \ s)^2=174.1 \ m$
The total elapsed time and distance traveled for the acceleration/deceleration phases are:
$t_{acc}+t_{dec}=23.99 \ s + 13.20 \ s = 37.19 \ s$ and
$(x-x_0)_{acc}+(x-x_0)_{dec}=316.5 \ m + 174.1 \ m = 491 \ m$
(a) If the stations are spaced 3.0 km = 3000 m apart, then there is a total of $\frac{15,000 \ m}{3,000 \ m}=5$ interstation segments.
A train making the entire trip would thus have a total of 5 interstation segments and 4 stops of 22 s each at the intermediate stations.
Since 491 m is traveled during acceleration and deceleration, 3000 m − 491 m = 2509 m of each segment is traveled at an average speed of $\overline{v}$ = 26.39 m/s.
The time for that 2509 m is given by $\Delta x = \overline{v}t$, so
$t_{constant \ speed}=\frac{\Delta x}{\overline{v}}=\frac{2509 \ m}{26.39 \ m/s}=95.07 \ s$
Thus a total interstation segment will take
37.19 s + 95.07 s = 132.26 s.
With 5 interstation segments of 132.26 s each, and 4 stops of 22 s each, the total time is given by
$t_{3.0 \ km}$ = 5(132.26 s) + 4(22 s) = 749 s = 12.5 min
(b) If the stations are spaced 5.0 km = 5000 m apart, then there is a total of $\frac{15,000 \ m}{5,000 \ m}=3$ interstation segments.
A train making the entire trip would thus have a total of 5 interstation segments and 2 stops of 22 s each at the intermediate stations.
Since 491 m is traveled during acceleration and deceleration, 5000 m − 491 m = 4509 m of each segment is traveled at an average speed of $\overline{v}$ = 26.39 m/s.
The time for that 4509 m is given by $\Delta x = \overline{v}t$, so
$t_{constant \ speed}=\frac{\Delta x}{\overline{v}}=\frac{4509 \ m}{26.39 \ m/s}=170.86 \ s$
Thus a total interstation segment will take
37.19 s + 170.86 s = 208.05 s.
With 3 interstation segments of 208.05 s each, and 2 stops of 22 s each, the total time is given by
$t_{5.0 \ km}$ = 3(208.05 s) + 2(22 s) = 668 s = 11.1 min
Q#67
A person driving her car at 35 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 28 m away from the near side of the intersection (Fig. 2–49). Should she try to stop, or should she speed up to cross the intersection before the light turns red? The intersection is 15 m wide. Her car’s maximum deceleration is $-5.80 \ m/s^2$ whereas it can accelerate from 45 km/h to 65 km/h in 6.0 s. Ignore the length of her car and her reaction time.
Answer:
The car’s initial speed is
$v_0=(35 \ km/h)\left(\frac{1000 \ m/km}{3600 \ s/h}\right)=9.722 \ m/s$
Case I: trying to stop.
The constraint is, with the braking deceleration of the car (a = −5.8 $m/s^2$), can the car stop in a 28-m displacement? The 2.0 seconds has no relation to this part of the problem.
Using Eq. $v^2=v_0^2+2a(x-x_0)$, the distance traveled during braking is as follows:
$x-x_0=\frac{v^2-v_0^2}{2a}$
$x-x_0=\frac{0-(9.722 \ m/s)^2}{2(-5.8 \ m/s^2)}=8.14 \ m$, She can stop the car in time.
Case II: crossing the intersection.
The constraint is, with the acceleration of the car
$a = \frac{\Delta v}{\Delta t}=\frac{65 \ km/h - 45 \ km/h}{6.0 \ s}=3.33 \ km/h/s$ or
$a = (3.33 \ km/h/s)\left(\frac{1000 \ m/km}{3600 \ s^2/h}\right)=0.9259 \ m/s^2$
can she get through the intersection (travel 43 m) in the 2.0 seconds before the light turns red?
Using Eq. $x-x_0=v_0t+\frac{1}{2}at^2$, the distance traveled during the 2.0 s is
$x-x_0=(9.722 \ m/s)(2.0 \ s)+\frac{1}{2}(0.9259 \ m/s^2)(2.0 \ s)^2=21.3 \ m$, She should stop.
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