Problem#1
A jet plane lands with a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s2 as it comes to rest. (a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest? (b) Can this plane land on a small tropical island airport where the runway is 0.800 km long?Answer:
(a) the minimum time interval needed before it can come to rest
We use
v = v0 + at
0 = 100 m/s – (5.00 m/s2)t
t = 20.0 s
and
v2 = v02 + 2a∆x
0 = (100 m/s)2 + 2(–5.00 m/s2)(∆x)
∆x = 1000 m
(b) At this acceleration the plane would overshoot the runway: No
Problem#2
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.60 m/s2 for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strike the tree?
Answer:
In the simultaneous equations:
vx = v0x + axt = v0x + (–5.60 m/s2)(4.20 s)
then, from
x = x0 + ½ (v0x + vx)t
62.4 m = 0 + ½ [vx + (5.60 m/s2)(4.20 s) + vx]
14.9 m/s = vx + 11.76 m/s
vx = 3.10 m/s
Problem#3
Within a complex machine such as a robotic assembly line, suppose that one particular part glides along a straight track. A control system measures the average velocity of the part during each successive interval of time ∆t0 = t0 – 0, compares it with the value vc it should be, and switches a servo motor on and off to give the part a correcting pulse of acceleration. The pulse consists of a constant acceleration am applied for time interval ∆tm = tm – 0within the next control time interval ∆t0. As shown in Fig. 1, the part may be modeled as having zero acceleration when the motor is off (between tm and t0). A computer in the control system chooses the size of the acceleration so that the final velocity of the part will have the correct value vc . Assume the part is initially at rest and is to have instantaneous velocity vc at time t0. (a) Find the required value of am in terms of vc and tm. (b) Show that the displacement ∆x of the part during the time interval ∆t0 is given by ∆x = vc(t0 – 0.5tm). For specified values of vc and t0, (c) what is the minimum displacement of the part? (d) What is the maximum displacement of the part? (e) Are both the minimum and maximum displacements physically attainable?
 |
| Fig.1 |
Answer:
(a) Along the time axis of the graph shown, let i = 0 and f = tm. Then
vx = v0x + axt
gives
vc = 0 + amtm
am = vc/tm
(b) The displacement between 0 and tm is
x – x0 = v0xt + ½ axt2 = 0 + ½ (vc/tm)tm2
x – x0 = ½ vctm
The displacement between tm and t0 is
x – x0 = v0xt + ½ axt2
x – x0 = vc(t0 – tm) + 0
The total displacement is
∆x = ½ vctm + vc(t0 – tm)
∆x = vc[t0 – ½tm]
(c) For constant vc and t0 , ∆x is minimized by maximizing tm to tm = t0. Then
∆xmin = vc[t0 – ½t0] = ½ vct0
This is realized by having the servo motor on all the time.
(d) We maximize ∆x by letting tm approach zero. In the limit
∆x = vc(t0 – 0) = vct0
(e) This cannot be attained because the acceleration must be finite
Problem#4
A ball starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 15.0 m, it comes to rest. (a) What is the speed of the ball at the bottom of the first plane? (b) How long does it take to roll down the first plane? (c) What is the acceleration along the second plane? (d) What is the ball’s speed 8.00 m along the second plane?
Answer:
(a) Take initial and final points at top and bottom of the incline. If the ball starts from rest,
v0 = 0, a = 0.500 m/s2, x – x0 = 9.00 m.
Then
v2 = v02 + 2a∆x
v2 = 0 + 2(0.500 m/s2)(9.00 m)
v = 3.00 m/s
(b) we use x – x0 = v0xt + ½ axt2
9.00 m = 0 + ½ (0.500 m/s2)t2
t = 6.00 s
(c) Take initial and final points at the bottom of the planes and the top of the second plane, respectively:
v0 = 3.00 m/s, v = 0, x – x0 = 15.00 m
then
v2 = v02 + 2a∆x
0 = (3.00 m/s)2 + 2a(15.00 m)
a = –3.00 m/s2
(d) Take the initial point at the bottom of the planes and the final point 8.00 m along the second:
v0 = 3.00 m/s, ∆x = 8.00 m and a = –3.00 m/s2, then
v2 = v02 + 2a∆x
v2 = (3.00 m/s)2 + 2(–3.00 m/s2)(8.00 m)
v = 2.05 m/s
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