Problem#1
A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration.Answer:
(a) We use
x – x0 = ½ (v + v0)t
becomes
40.0 m – 0 = ½ (v0 + 2.80 m/s)(8.50 s)
which yields
v0 = 6.61 m/s
(b) the acceleration is
v = v0 + at
2.8 m/s = 6.61 m/s + a(8.85 s)
which yields
a = –0.448 m/s2
Problem#2
A 745i BMW car can brake to a stop in a distance of 121 ft. from a speed of 60.0 mi/h. To brake to a stop from a speed of 80.0 mi/h requires a stopping distance of 211 ft. What is the average braking acceleration for (a) 60 mi/h to rest, (b) 80 mi/h to rest, (c) 80 mi/h to 60 mi/h? Express the answers in mi/h/s and in m/s2.
Answer:
(a) Let i be the state of moving at 60 mi/h and f be at rest
We use
vxf2 = vxi2 + 2ax(xf – xi)
0 = (60 mi/h)2 + 2ax(121 ft – 0)(1 mi/5280 ft)
which yields
ax = –(3600 mi/242 h2)(5280 ft/1 mi)(1 h/3600 s) = –21.8 mi/h.s
or
ax = (–21.8 mi/h.s)(1609 m/1 mi)(1 h/3600 s) = –9.75 m/s2
(b) Similarly,
0 = (80 mi/h)2 + 2ax(211 ft – 0)(1 mi/5280 ft)
which yields
ax = –(6400 mi/422 h2)(5280 ft/1 mi)(1 h/3600 s) = –22.2 mi/h.s
or
ax = (–22.2 mi/h.s)(1609 m/1 mi)(1 h/3600 s) = –9.94 m/s2
(c) Let i be moving at 80 mi/h and f be moving at 60 mi/h.
vxf2 = vxi2 + 2ax(xf – xi)
(60 mi/h)2 = (80 mi/h)2 + 2ax(211 ft – 121 ft)(1 mi/5280 ft)
which yields
ax = –(2800 mi/90 h2)(5280 ft/1 mi)(1 h/3600 s) = –22.8 mi/h.s
or
ax = (–22.8 mi/h.s)(1609 m/1 mi)(1 h/3600 s) = –10.2 m/s2
Problem#3
Figure 1 represents part of the performance data of a car owned by a proud physics student. (a) Calculate from the graph the total distance traveled. (b) What distance does the car travel between the times t = 10 s and t = 40 s? (c) Draw a graph of its acceleration versus time between t = 0 and t = 50 s. (d) Write an equation for x as a function of time for each phase of the motion, represented by (i) 0a, (ii) ab, (iii) bc. (e) What is the average velocity of the car between t = 0 and t = 50 s?
 |
| Fig.1 |
(a) Total displacement = area under the v vs t , curve from t = 0 to 50 s.
∆x = ½ (50 m/s)(15 s) + (50 m/s)(40 – 15) s + ½ (50 m/s)(10 s)
∆x = 1875 m
(b) From t =10 s to t = 40 s, displacement is
∆x = ½ (50 m/s + 33 m/s)(5 s) + (50 m/s)(25 s) = 1457 m
(c) For 0 ≤ t ≤ 15 s: a1 = ∆v/∆t = (50 m/s – 0)/(15 s – 0) = 3.3 m/s2
For 15 s ≤ t ≤ 40 s: a2 = ∆v/∆t = 0
For 40 ≤ t ≤ 50 s: a3 = ∆v/∆t = (0 – 50 m/s)/(50 s – 40 s) = –5.0 m/s2
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| Fig.2 |
(d) equation for x as a function of time for each phase of the motion, represented by
(i) 0a; x1 = 0 + ½ a1t2 = ½ (3.3 m/s2)t2 or x1 = (1.67 m/s2)t2
(ii) ab; x2 = ½ (15 s)(50 m/s – 0) + (50 m/s)(t – 15 s) or x2 = (50 m/s)t – 375 m
(iii) bc; for 40 s ≤ t ≤ 50 s,
x3 = 375 m + 1250 m + ½ (–5.0 m/s2)(t – 40 s)2 + (50 m/s)(t – 40 s)
x3 = (250 m/s)t – (2.50 m/s2)t2 – 4375 m
(e) the average velocity of the car between t = 0 and t = 50 s is
vavg = ∆x/∆t = total displacement/total time
vavg = 1875 m/50 s = 37.5 m/s
Problem#4
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.5 m/s. The driver of the Thunderbird realizes he must make a pit stop, and he smoothly slows to a stop over a distance of 250 m. He spends 5.00 s in the pit and then accelerates out, reaching his previous speed of 71.5 m/s after a distance of 350 m. At this point, how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed?
Answer:
The time for the Ford to slow down we find from
x = x0 + ½(v + v0)t
t = 2∆x/(v + v0)
t = 2(250 m)/(71.5 m/s + 0) = 6.99 s
Its time to speed up is similarly
t = 2(350 m)/(71.5 m/s + 0) = 9.79 s
The whole time it is moving at less than maximum speed is
6.99 s + 5.00 s + 9.79 s = 21.8 s
The Mercedes travels
x = x0 + ½(v + v0)t
x = 0 + ½ (71.5 m/s + 71.5 m/s)(21.8 s) = 1558 m
while the Ford travels 250 m + 350 m = 600 m, to fall behind by 1558 m – 600 m = 958 m.
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