1.
Δv=-8m/s-12m/s=20m/s,t=10s,所以a=Δv/t=-2m/s^2,那么F=ma=2kg*(-2m/s^2)=-4N.所以所加的力大小为4N,方向为x轴负方向
2.
S=v0t+1/2at^2=12m/s*10s+1/2*(-2m/s^2)(10s)^2=20m
即10s后距D20米
3.
正向最大距离时物体速度为0,那么t1=(0-12m/s)/(-2m/s^2)=6s
此时Sm=v0t1+1/2at1^2=36m
现在初速度为v0'=2*12m/s=24m/s,那么t1'=-v0'/a'
Sm=v0't1'+1/2a't1'^2=-576m^2/s^2/a'+288m^2/s^2/a'=-288m^2/s^2/a'=36m
a'=-8m/s^2,所需的力F'=ma'=-16N,是原来的4倍
t1'=-24m/s/(-8m/s^2)=3s,是原来的1/2