識幾多答幾多啦，感激萬分! ^^＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊Useful equations of uniformly accelerated motions (Linear) s = 1/2 (u + v) tv = u + a ts = u t + 1/ 2 a t²v2 = u2 + 2 a s s = Displacement / u = Initial Velocity / v = Final Velocity / a = Acceleration / t = Time＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊
8) A car takes 247.5 seconds to travel between two stations A and B, 3km apart, starting from rest and finishing at rest. The acceleration is uniform for 45 seconds and the deceleration is also uniform for the last 30 seconds, the velocity being constant for the remaining time. Determine
a) this velocity in km hr -1,
b) the acceleration in ms-2,
c) the deceleration in ms-2
9. A car A starts from rest with a uniform acceleration of 0.6ms-2. A second car B starts from the same point, 4 second late and follows the same path with an acceleration of 0.9 ms-2. How far will the cars have traveled when B passes A?
10) Before taking off, an aircraft traveling with a constant acceleration makes a run on a field of
1800m in 12 seconds from rest. Determine
a) the acceleration,b) the velocity with which it takes off,
c) the distance covered in the 1st and the 12th seconds.
十萬火急MECHANIC -KINETICS 動力學 (B)
2011-04-14 4:43 am
回答 (1)
2011-04-14 10:15 am
✔ 最佳答案
Question 8: The trip is divided into 3 parts.Part 1: under acceleration,
Time for acceleration = 45s (given). Distance covered = s1
Part 2 under constant velocity
time for constant velocity = 247.5 s – 45s – 30s = 172.5 s, Distance covered = s2
Part 3 under deceleration
Time for deceleration = 30s (given). Distance covered = s3
Let v = velocity that is attainsed after acceleration. That is also the constant velocity of the car the car in part 2.
Part 1 distance covered s1 = average velocity x time = (1/2v)(45)
Part 2 distance covered s2 = constant velocity x time = (v)( 172.5)
Part 3 distance covered s3 = average velocity x time = (1/2v)(30)
Total distance covered = 3 km = 3000 m = s1 + s2 + s3
3000 m = (1/2v)(45) + (v)(172.5) + (1/2v)(30)
22.5 v + 172.5 v + 15v = 3000
210v = 3000
v = 3000/210 = 14.2857 m/s =(14.2857/1000)(600) = 51.4286 km/h
a) velocity is 51.4286 km/h
b) a = (v – u)/t = (14.2857 m/s – 0 m/s)/45s = 0.3175 m/s^2
acceleration = 0.3175 m/s^2
c) a = (v – u)/t = (0 m/s - 14.2857 m/s )/30s = - 0.4762 m/s^2
deceleration = 0.4762 m/s^2
Question 9.
Let t be the time elapsed before car B caught car A.(that is the time car A had travelled)
Since car b started traveled 4 s later, the time car b had traveled = t - 4
The distance car A traveled = distance car B traveled
ut + (½)(a of car A)t^2 = u(t-4) + (1/2)(a of car B)(t-4)^2
Since u = 0 m/s
(½)(0.6)t^2 = (1/2)(0.9)(t-4)^2
0.3t^2 = 0.45 (t – 4)^2
t^2 -24t + 48 = 0
Solve for t, Using quadratic formula
t = 24 +[24^2 –(4)(48) ]^0.5/2
t = 21.798 s
The distance traveled s = (½)(0.6)(21.798)^2 = 142.54 m
The cars have traveled 142.54 m
Question 10
(a) u = 0m/s, s = 1800 m,t = 4s
s = ut +1/2 at^2
1800 = 0 +1/2a(4)^2
3600 = 8a
a = 3600/8
a = 450 m/s^2
(b) u = 0 m/s, v = ?, a = 450 m/s^2, t = 4s
450 =( v – 0)/4
v = (450)(4) = 1800 m/s
(c) distance covered 1 s = 1/2 a t^2 = ½(450)(1)^2 = 225 m
distance covered after 12 s before take-off = 1800 m
distance covered between 1st s and 12th second = 1800 m – 225 m = 1575 m
收錄日期: 2021-04-29 17:05:04
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