Problem#1
In Fig. 1, a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 220 m. If the red car has a constant velocity of 20 km/h, the cars pass each other at x = 44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity and (b) the constant acceleration of the green car?
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| Fig.1 |
Answer:
Let d be the 220 m distance between the cars at t = 0, and v1 be the 20 km/h = 50/9 m/s speed (corresponding to a passing point of x1 = 44.5 m) and v2 be the 40 km/h =100/9 m/s speed (corresponding to a passing point of x2 = 76.6 m) of the red car. We have two equations (from eq. x – x0 = v0t + ½ at2)
d – x1 = v0t1 + ½ at12 where t1 = x1/v1
d – x2 = v0t2 + ½ at22 where t2 = x2/v2
We simultaneously solve these equations and obtain the following results:
(a) The initial velocity of the green car is vo = − 13.9 m/s. or roughly − 50 km/h (the negative sign means that it’s along the –x direction).
(b) The corresponding acceleration of the car is a = − 2.0 m/s2 (the negative sign means that it’s along the –x direction).
Problem#2
A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first of that distance, its acceleration is +2.25 m/s2. Through the rest of that distance, its acceleration is –0.750 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed? (c) Graph position x, velocity v, and acceleration a versus time t for the trip.
Answer:
(a) Equation x – x0 = v0t + ½ at2 is used for part 1 of the trip
∆x1 = v01t1 + ½ a1t12
Where a1 = 2.25 m/s2 and ∆x1 = 900/4 m = 225 m
and Eq. x – x0 = vt – ½at2 is used for part 2:
∆x2 = v2t2 – ½ a2t22
Where a2 = –0.750 m/s2 and ∆x1 = ¾ (900 m) = 675 m
In addition, vo1 = v2 = 0.
Solving these equations for the times and adding the results gives
t = t1 + t2 = 56.6 s.
(b) Equation
v2 = v012 + 2a1∆x1
is used for part 1 of the trip:
v2 = 0 + 2(2.25)(225 m)
v = 31.8 m/s for the maximum speed.
Problem#3
(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s2 and subway stations are located 806 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 20 s at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph x, v, and a versus t for the interval from one start-up to the next.
Answer:
We assume the train accelerates from rest (v0 = 0 and x0 = 0 ) at a1 = +134 m/s2 until the midway point and then decelerates at a2 = −134 m/s2 until it comes to a stop (v2 = 0) at the next station.
The velocity at the midpoint is v1, which occurs at x1 = 806/2 = 403m.
(a) Equation
v12 = v012 + 2a1∆x1
leads to
v12 = 0 + 2(134 m/s2)(403 m)
v1 = 32.9 m/s
(b) The time t1 for the accelerating stage is
x1 = v01t1 + ½ a1t12
403 m = 0 + ½ (1.34 m/s2)t12
t1 = 24.53 s
Since the time interval for the decelerating stage turns out to be the same, we double this result and obtain t = 49.1 s for the travel time between stations
(c) With a “dead time” of 20 s, we have T = t + 20 = 69.1 s for the total time between start-ups. Thus, Eq. vavg = ∆x/∆t gives
vavg = 806 m/69.1 s = 11.7 m/s
(d) The graphs for x, v and a as a function of t are shown below (Fig.2a, Fig2b and Fig2c). The third graph, a(t), consists of three horizontal “steps” — one at 1.34 m/s2 during 0 < t < 24.53 s, and the next at –1.34 m/s2 during 24.53 s < t < 49.1 s and the last at zero during the “dead time” 49.1 s < t < 69.1 s).
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| Fig.2 |
Problem#4
You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of v0 = 55 km/h; your best deceleration rate has the magnitude a = 5.18 m/s2. Your best reaction time to begin braking is T = 0.75 s. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at 55 km/h if the distance to the intersection and the duration of the yellow light are (a) 40 m and 2.8 s, and (b) 32 m and 1.8 s? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).
Answer:
We take the direction of motion as +x, so a = –5.18 m/s2 , and we use SI units, so v0 = 55(1000/3600) = 15.28 m/s.
(a) The velocity is constant during the reaction time T, so the distance traveled during it is
dr = v0T – (15.28 m/s) (0.75 s) = 11.46 m.
We use
v2 = v02 + 2a∆x
(with v = 0) to find the distance db traveled during braking:
0 = (15.28 m/s)2 + 2(–5.18 m/s2)∆x
∆x = 22.53 m
Thus, the total distance is
dr + db = 34.0 m
which means that the driver is able to stop in time
And if the driver were to continue at v0, the car would enter the intersection in
t = (40 m)/(15.28 m/s) = 2.6 s
which is (barely) enough time to enter the intersection before the light turns, which many people would consider an acceptable situation.
(b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than the distance to the intersection, so the driver cannot stop without the front end of the car being a couple of meters into the intersection. And the time to reach it at constant speed is
32/15.28 = 2.1 s
which is too long (the light turns in 1.8 s).
The driver is caught between a rock and a hard place.
Problem#5
You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0 s (long enough for you to look at the phone and yell, “I won’t do that!”). At the beginning of that 2.0 s, the police officer begins braking suddenly at 5.0 m/s2. (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another 0.40 s to realize your danger and begin braking. (b) If you too brake at 5.0 m/s2, what is your speed when you hit the police car?
Answer:
(a) Note that 110 km/h is equivalent to 30.56 m/s. During a two-second interval, you travel 61.11 m.
The decelerating police car travels (using Eq. x – x0 = v0t + ½ at2) 51.11 m. In light of the fact that the initial “gap” between cars was 25 m, this means the gap has narrowed by 10.0 m – that is, to a distance of 15.0 m between cars.
(b) First, we add 0.4 s to the considerations of part (a). During a 2.4 s interval, you travel 73.33 m. The decelerating police car travels (using Eq. x – x0 = v0t + ½ at2) 58.93 m during that time. The initial distance between cars of 25 m has therefore narrowed by 14.4 m. Thus, at the start of your braking (call it t0) the gap between the cars is 10.6 m. The speed of the police car at t0 is
30.56 – 5(2.4) = 18.56 m/s
Collision occurs at time t when
xyou = xpolice (we choose coordinates such that your position is x = 0 and the police car’s position is x = 10.6 m at t0)
for each car:
xpolice – 10.6 = 18.56(t – t0) – ½ (5)(t – t0)2
xyou = 30.56(t – t0) – ½ (5)(t – t0)2
Subtracting equations, we find
10.6 = (30.56 – 18.56)(t − t0)
⇒ t − t0 = 0.883 s
At that time your speed is
30.56 + a(t − t0) = 30.56 – 5(0.883)
≈ 26 m/s (or 94 km/h).
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