2002CAP Prize Exam加拿大高中物理大奖赛真题答案下载

历年CAP High School Prize Exam加拿大高中物理大奖赛

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2002 CAP Prize Exam完整版真题免费下载

共计3小时考试时间

此套试卷由25道选择题以及3道大题组成

每道大题含有不同数量的小题

2002CAP Prize Exam加拿大物理奥林匹克完整版答案免费下载

部分真题答案预览：

Part A选择题部分：

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

Part B简答题部分

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 final 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 pass­ing 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 final 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 manœuvre. With vcar = 111 km/hr and (hope­fully!) 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 final rela­tive speed which could be as large as 20 km/hr , giving ta =4.0 × 20/11 = 7.3 s. Taking into account her re­action time during which she travels at 111 km/hr, the distance she covers over the whole passing manœuvre, plus the safety margin at the end, is now

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