计算一道英文数学题 关于velocity速度的
Atatimeoftseconds,aparticlemovesadistanceofsmetersfromitsstartingpoint,wheres=5sin(1....
At a time of t seconds, a particle moves a distance of s meters from its starting point, where s=5 sin(1.7 t)
a) Find the average velocity (correct to four decimal places) between t=1 and t=1+h for the following values of h.
If h=0.1 then the average velocity over the time period [1,1.1] equal__________________m/s.
If h=0.01 then the average velocity over the time period [1,1.01] equals______________m/s.
If h=0.001 then the average velocity over the time period [1,1.001] equals___________________m/s.
If h=0.0001 then the average velocity over the time period [1,1.0001] equals_____________________m/s.
(b) Using the results of part (a), estimate the instantaneous velocity at time t=1. 展开
a) Find the average velocity (correct to four decimal places) between t=1 and t=1+h for the following values of h.
If h=0.1 then the average velocity over the time period [1,1.1] equal__________________m/s.
If h=0.01 then the average velocity over the time period [1,1.01] equals______________m/s.
If h=0.001 then the average velocity over the time period [1,1.001] equals___________________m/s.
If h=0.0001 then the average velocity over the time period [1,1.0001] equals_____________________m/s.
(b) Using the results of part (a), estimate the instantaneous velocity at time t=1. 展开
2个回答
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Answer:
a) For h=0.1, the average velocity is [S(1.1)-S(1)]/(1.1-1) = -1.8045 (m/s); similarly, for h=0.01, h=0.001, and h=0.0001, the velocities are, respectively, -1.167 (m/2), -1.1025 (m/s) and -1.0955 (m/s).
b) The case for h=0.0001 is a good approximation of the instantaneous velocity at t=1, i.e., -1.0955 (m/s).
Actually, the velocity at any instant t is V(t) = dS(t)/dt = 5x1.7cos(1.7t). Thus V(1) = 5x1.7 cos(1.7x1) = -1.0952 (m/s). This is very close to the case h=0.0001.
a) For h=0.1, the average velocity is [S(1.1)-S(1)]/(1.1-1) = -1.8045 (m/s); similarly, for h=0.01, h=0.001, and h=0.0001, the velocities are, respectively, -1.167 (m/2), -1.1025 (m/s) and -1.0955 (m/s).
b) The case for h=0.0001 is a good approximation of the instantaneous velocity at t=1, i.e., -1.0955 (m/s).
Actually, the velocity at any instant t is V(t) = dS(t)/dt = 5x1.7cos(1.7t). Thus V(1) = 5x1.7 cos(1.7x1) = -1.0952 (m/s). This is very close to the case h=0.0001.
2014-11-01
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躲进了洞穴哈哈
在山脚的洞穴被塞满残枝败叶后
迁移到了山顶的洞穴
谁也不敢迎风直立行走
只有重新用四肢爬行
才能躲开风头
但还是有太多的人
在山脚的洞穴被塞满残枝败叶后
迁移到了山顶的洞穴
谁也不敢迎风直立行走
只有重新用四肢爬行
才能躲开风头
但还是有太多的人
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