2002CAP Prize Exam加拿大高中物理大獎賽真題答案下載

歷年CAP High School Prize Exam加拿大高中物理大獎賽

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2002 CAP Prize Exam完整版真題免費下載

共計3小時考試時間

此套試卷由25道選擇題以及3道大題組成

每道大題含有不同數量的小題

2002CAP Prize Exam加拿大物理奧林匹克完整版答案免費下載

部分真題答案預覽：

Part A選擇題部分：

1. C
2. B
3. C
4. D
5. A

Part B簡答題部分

Solution to Problem 1：

(a) It is convenient—but not essential to answer this—to work in the rest-frame of the slow car. Then we only have to deal with the relative ?nal velocity v of the cars, and their relative acceleration a, the latter being also the acceleration of the passing car with respect to the ground. We are interested in the position of the front of the pass-ing car with respect to the front of the slow car. Let ta be the time it takes to reach speed v at acceleration a = v/ta. Furthermore, let tb be the time it takes the front of the passing car to reach its ?nal position, four car lengths ahead of the slow car, at speed v. Calling T = ta + tb the total time it takes to pass the slow car, we can write for the position x of the front of the passing car:

1 ₂

x(T )= ?4.0L + at+ vtb

a

2 =4.0L

where L =4.2 m is the length of a car. Eliminating a = v/ta and solving for tb yields

8.0L 1

tb = ? ta

v 2

so that

8.0L 1

T =+ ta

v 2 With v = 11 km/hr (3.06

m/s) and ta =4.0 s, we obtain T = 13 s.

(b) The distance d that includes the distance the passing car covers after its driver notices the truck, plus the safety margin at the end, is

d =(vcar + vtruck)(tb +2.0 s)

 

8.0L 1

=(vcar + vtruck)? ta +2.0s

v 2

where the 2.0 s term accounts for the safety margin at the end of the man?uvre. With vcar = 111 km/hr and (hope-fully!) vtruck = 90 km/hr (25 m/s) at the most, we obtain d = 614 m, which is greater than the initial distance of taken.

Does the driver have time to pull in behind the slow car, given that braking would risk a collision with the car right behind her and that she must keep at a distance of at least

2.0 m behind the slow car if she pulls in? Well, when she sees the truck (at time ta =4.0 s, she is at a distance 3.0L ? vta/2=6.5 m behind the slow car. But during the 1.6 s it takes her to react, she has covered a further

4.9 m relative to the front car, so she is only about 1.6 m behind and would have to brake to keep at a safe distance from the car in front.

Her other alternative is to keep accelerating at the same rate. Now ta =4.0v^f/v, where v^f is her new ?nal rela-tive speed which could be as large as 20 km/hr , giving ta =4.0 × 20/11 = 7.3 s. Taking into account her re-action time during which she travels at 111 km/hr, the distance she covers over the whole passing man?uvre, plus the safety margin at the end, is now

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