A sudent is running on a straight road to try catching a bus ahead. The student is running at a speed of 5ms-1 and the bus accelerates frm rest. The distance between the student and the front door of the is 10m. Suppose the student keeps running at the same speed and the bus accelerate with a uniform acceleration of 0.8ms-2.
Question:
1)how long does it take for the student to reach the front door of the bus? answer:2.5s
2)At the moment the student reaches the front door of the bus, the driver applies the brake and the bus decelerates at a rate of 1ms-2. Find the distance further travelled by the bus before it stops. answer:1m
中四physics velocity (我要列式)
2009-04-19 5:40 am
回答 (2)
2009-04-19 7:49 am
✔ 最佳答案
1)Consider the bus:
u = 0 m s-1, ..a = 0.8 m s-1
s = ut + (1/2)at2
s = (0)t + (1/2)(0.8)t2
s = 0.4t2
Consider the student:
speed = 5 m s-1
distance run in m = 10 + 0.4t2 = 5t
0.4t2 - 5t + 10 = 0
2t2 - 25t + 50 = 0
(2t - 5)(t - 10) = 0
t = 2.5 s or t = 10 s (rejected)
(t = 10 s is rejected. This is the time used for the student to overtakes the bus and then the bus overtakes the student.)
Time taken = 2.5 s
2)
Consider the acceleration of the bus:
u = 0 m s-1, ..a = 0.8 m s-1, ..t = 2.5 s
v = u + at
v = 0 + (0.8)(2.5)
v = 2 m s-1
Consider the deceleration of the bus:
u = 2 m s-1, ..a = -1 m s-2, ..v = 0
v2 = u2 + 2as
(0)2 = (2)2 + 2(-1)s
s = 2 m
(The given answer is wrong. If the deceleration is 2 m s-2, the answer is 1 m.)
=
2009-04-19 7:27 am
The answer for 2) is really 1 m?
收錄日期: 2021-04-29 18:16:57
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