Problem#1
Cars A and B move in the same direction in adjacent lanes. The position x of car A is given in Fig. 1, from time t = 0 to t = 7.0 s. The figure’s vertical scaling is set by xs = 32.0 m. At t = 0, car B is at x = 0, with a velocity of 12 m/s and a negative constant acceleration aB. (a) What must aB be such that the cars are (momentarily) side by side (momentarily at the same value of x) at t = 4.0 s? (b) For that value of aB, how many times are the cars side by side? (c) Sketch the position x of car B versus time t on Fig. 1. How many times will the cars be side by side if the magnitude of acceleration aB is (d) more than and (e) less than the answer to part (a)? |
| Fig.1 |
(a) We note that
vA = 12/6 = 2 m/s (with two significant figures understood)
Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s
This must be the value of x for car B at that time; we use:
x – x0 = v0Bt + ½ at2
28 – 0 = (12 m/s)t + ½ aBt2
Where t = 4.0 s
This yields aB = –2.5 m/s2
(b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB? The requirement is
20 + 2t = 12t + ½ aBt2
20 + 2t = 12t + ½ (–2.5)t2
t2 – 8t + 16 = 0
There are two distinct roots unless the discriminant
D = (b2 – 4ac)1/2 = [82 – 4(16)]1/2 = 0
In our case, it is zero – which means there is only one root. The cars are side by side only once at t = 4 s.
(c) A sketch is shown below (Fig.2). It consists of a straight line (xA) tangent to a parabola (xB) at t = 4.
 |
| Fig.2 |
(d) We only care about real roots, which means
102 − 2(−20)(aB) ≥ 0
If |aB| > 5/2 then there are no (real) solutions to the equation; the cars are never side by side.
(e) Here we have
102 − 2(−20)(aB) > 0 ⇒ two real roots
The cars are side by side at two different times.
Problem#2
As two trains move along a track, their conductors suddenly notice that they are headed toward each other.Figure 2 gives their velocities v as functions of time t as the conductors slow the trains. The figure’s vertical scaling is set by vs = 40.0 m/s. The slowing processes begin when the trains are 200 m apart. What is their separation when both trains have stopped?
 |
| Fig.2 |
The displacement (Δx) for each train is the “area” in the graph (since the displacement is the integral of the velocity).
Each area is triangular, and the area of a triangle is
1/2(base) × (height)
Thus, the (absolute value of the) displacement for one train
(1/2)(40 m/s)(5 s) = 100 m,
and that of the other train is
(1/2)(30 m/s)(4 s) = 60 m
The initial “gap” between the trains was 200 m, and according to our displacement computations, the gap has narrowed by 160 m. Thus, the answer is
200 – 160 = 40 m.
Problem#3
When a high-speed passenger train traveling at 161 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D = 676 m ahead (Fig. 3). The locomotive is moving at 29.0 km/h. The engineer of the high-speed train immediately applies the brakes. (a) What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided? (b) Assume that the engineer is at x = 0 when, at t = 0, he first spots the locomotive. Sketch x(t) curves for the locomotive and highspeed train for the cases in which a collision is just avoided and is not quite avoided.
 |
| Fig.3 |
In this solution we elect to wait until the last step to convert to SI units. We start with Eq.
x – x0 = ½ (v + v0)t
and denote the train’s initial velocity as vt and the locomotive’s velocity as vA (which is also the final velocity of the train, if the rear-end collision is barely avoided). We note that the distance Δx consists of the original gap between them, D, as well as the forward distance traveled during this time by the locomotive vlt. Therefore,
(vt + vl)/2 = ∆x/t = (D + vlt)/t = D/t + vl
We now use Eq.
v = v0 + at
to eliminate time from the equation. Thus,
(vt + vl)/2 = D/[(vl – vt)/a] + vl
which leads to
a = [(vt + vl)/2 – vl][(vl – vt)/D]
a = –(vl – vt)2/2D
hence,
a = –[(29 km/h)2 – (161 km/h)]2/(2 x 0.676 km) = –12888 km/h2
which we convert as follows:
a = [–12888 km/h2][1000 m/1km][1h/3600 s]2 = –0.994 m/s2
so that its magnitude is |a| = 0.994 m/s2
A graph is shown here for the case where a collision is just avoided (x along the vertical axis is in meters and t along the horizontal axis is in seconds). The top (straight) line shows the motion of the locomotive and the bottom curve shows the motion of the passenger train. The other case (where the collision is not quite avoided) would be similar except that the slope of the bottom curve would be greater than that of the top line at the point where they meet.
Post a Comment for "Constant Acceleration Problems and Solutions 3"