物理答案_2024-2025高一（7-7月题库）_2024年11月试卷_1121山西省太原市2024-2025学年高一上学期11月期中考试

2024 ~ 2025 学年第一学期高一年级期中学业诊断
物理参考答案及评分建议
一、单项选择题：本题包含10小题，每小题3分，共30分。
题号 1 2 3 4 5 6 7 8 9 10
答案 A D C C D B A A C C
二、多项选择题：本题包含5小题，每小题3分，共15分。
题号 11 12 13 14 15
答案 AC BD BC AC ACD
三、实验题：本题包含2小题，共14分。
16.（9分）
（1）B（2分） （2）1.0或0.99(3分） （3）1.9(4分）
17.（5分）
1
实验原理：查阅本地重力加速度的值 自由落体位移时间关系h  gt 2 (2分）
2
实验器材：毫米刻度尺（1分）
实验步骤：一名同学竖直握刻度尺上端，刻度尺下端零刻度线对齐被测试同学的虎口。静
止释放刻度尺，被测试同学握住刻度尺，被测试同学的虎口对应位置即反应时间内刻度尺
下落的高度。代入公式，计算被测试同学的反应时间。（2分）
（其余合理答案均可得分）
四、计算题：本题包含5小题，共41分。
18.（7分）
汽车通过该位置的瞬时速度可近似用平均速度表示
高一物理答案 第 1 页
{#{QQABDYQUogioABAAAQgCAQVACkCQkgGACSgGQFAAoAIByBFABAA=}#}v= ···························································································（4分）
∆ 

v= =25m/s=90km/h ································································（2分）

.  
90km/h>80km/h 汽车超速·······························································（1分）
19.（7分）
（1）运动员下落过程做自由落体运动
v2 -02gh···················································································（2分）
0
v  2gh 8m/s··········································································（1分）
0
（2）以竖直向下为正方向，加速度为
v-v
a  0 ······················································································· （2分）
Δt
a 18m/s2 ··················································································（1分）
加速度大小为18m/s2，方向竖直向上···················································（1分）
20.（8分）
（1）玩具小车做匀加速直线运动
1
x  at 2 ·····················································································（2分）
1 2 11
t 3s···························································································（1分）
1
（2）玩具小车匀加速至B点，在B点速度
v at 3m/s··············································································· （2分）
1 11
玩具小车接着保持加速后的最大速率3m/s沿BC转弯，最后匀减速至D点
0v 2
x  1 ···················································································（2分）
2 2a
2
x 2.25m··················································································· （1分）
2
21.（9分）
（1）计时开始后第1s内运动员位移
高一物理答案 第 2 页
{#{QQABDYQUogioABAAAQgCAQVACkCQkgGACSgGQFAAoAIByBFABAA=}#}1
x  at2 0.5m········································································（1分）
1 2
计时开始后第1s运动员运动的时间
t 1t ······················································································（1分）
0
计时开始后前两秒内运动员位移
1
x a

1t
2
··············································································（2分）
2
计时开始后运动员第2s内位移
x 4m
2
计时开始后前两秒内运动员位移
x x x 4.5m··········································································（1分）
1 2
t 0.5s························································································（1分）
0
（2）计时开始后第1s内，运动员的位移
1
x  at2 0.5m
1 2
t 1t 0.5s·············································································（1分）
0
运动员起跑的加速度的大小
a 4m/s2······················································································（2分）
22.（10分）
（1）根据速度时间关系
··············································································（1分）
m =   +    
A车加速到最大速度的时间为
=  
A车行驶2.5s后的位移为
+  m
=   +  m  −  
······················································································（2分）
=    高一物理答案 第 3 页
{#{QQABDYQUogioABAAAQgCAQVACkCQkgGACSgGQFAAoAIByBFABAA=}#}这段时间内B车的位移
··············································································（1分）
=    =   
此时A、B两车相距
·········································································（1分）
=   +   −  
······················································································（1分）
=   
（2）当A车减速到两车速度相等时恰好不相撞
··············································································（1分）
=  m−    
··········································································（2分）

m−  
−     =  
A车减速时的加速度的大小为
················································································（1分）
2
=  .  m/s
高一物理答案 第 4 页
{#{QQABDYQUogioABAAAQgCAQVACkCQkgGACSgGQFAAoAIByBFABAA=}#}