一辆汽车沿圆周以Vo=7.0 m/s 的初速率匀减速行驶,经过1s,汽车的加速度与速度之间的夹角θ1=135°,又经过3
1个回答

在 t =1s,(v0 - at*1)^2/R = at

由经过3s,即 t = 4s,(v0 - at*4)^2/R = at*tan30°

联立两式,解出

{R = 7.52102,at = 2.58808},或者 {R = 80.959,at = 0.518847}

经过1s,汽车的法向加速度

an(1) = (7.0 - 2.58808*1)^2/7.52102 = 2.588 m/s^2

或 an(1) = (7.0 - 0.518847*1)^2/ 80.959 = 0.518847 m/s^2