Question

质点在xy平面内的运动方程x=6t,y=4t^2-8(SI),则t=1s时,质点的切向加速度和法向加速度各是多少?

Answer 1

vx = 6

vy = 8t

ax = 0

ay = 8

t=1 时 v=(6,8),a=(0,8)

切向加速度是

v/|v|^2 (其中为v,a内积)

= (6*0+8*8)(6,8)/(6^2+8^2)

= (96/25,128/25)

法向加速度是

(0,8) - (96/25,128/25)

= (-96/25,72/25)