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15问答网 > 质点运动方程为x=3t^2-4t,y=-t^3+3t,求（1）质点的初速度,（2）t=1s时质点的加速度

质点运动方程为x=3t^2-4t,y=-t^3+3t,求（1）质点的初速度,（2）t=1s时质点的加速度

在线急等

2025-03-16 01:08:49

推荐回答（1个）

回答1：

(1)
x=3t^2-4t
dx/dt = 6t-4
y=-t^3+3t
dy/dt = -3t^2+3
质点的初速度 = ( dx/dt , dy/dt) = ( 6t-4 , -3t^2+3)
(2)
dx/dt = 6t-4
d^2x/dt^2 =6
dy/dt = -3t^2+3
d^2y/dt^2 = -6t
d^2y/dt^2 |t=1 = -6
t=1s时质点的加速度 = ( 6, -6)

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