1.the velocity of a particle moving back and forth on a line
1.the velocity of a particle moving back and forth on a line is v=ds/dt=6sin2t m/sec for all t.If s=0 when t=0,find the value of s when t=π/2 sec.
2.the acceleration of a particle moving back and forth on a line is a=d^2s/dt^2=π^2 × cosπt m/sec^2 for all t.If s =0 and v=8m/sec when t=0,find s when t=1 sec.
第二题s是等于1吗?
2.the acceleration of a particle moving back and forth on a line is a=d^2s/dt^2=π^2 × cosπt m/sec^2 for all t.If s =0 and v=8m/sec when t=0,find s when t=1 sec.
第二题s是等于1吗?
赞 守候阳光111 幼苗
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就是解微分方程,初值都给了
1 ds/dt=6sin2t直接分离变量
ds=3sin2t d(2t)积分得s=-3cos2t+C
t=0,s=0,得C=3,于是s=3-3cos2t
when t=π/2,s=3-3cosπ=6
2.d^2s/dt^2=π^2cosπt 直接积分两次
ds/dt=πsin(πt)+C1
s=-cos(πt)+C1t+C2
when t=0,v=8, => C1=8
when t=0,s=0, => s=-1+C2=0,=>C2=1
so: s=8t+1-cos(πt)
when t=1, s=8+1-cosπ=10
1 ds/dt=6sin2t直接分离变量
ds=3sin2t d(2t)积分得s=-3cos2t+C
t=0,s=0,得C=3,于是s=3-3cos2t
when t=π/2,s=3-3cosπ=6
2.d^2s/dt^2=π^2cosπt 直接积分两次
ds/dt=πsin(πt)+C1
s=-cos(πt)+C1t+C2
when t=0,v=8, => C1=8
when t=0,s=0, => s=-1+C2=0,=>C2=1
so: s=8t+1-cos(πt)
when t=1, s=8+1-cosπ=10
1年前
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