Solution.
v0=0m/s;
v=20m/s;
a=2a1;
t=120s;
a)
b)2av2+2a1v2+vτ=2000;
2av+a1v+τ=120;
3v2+4a1vτ=8000a1;
3v+2a1τ=240a1;
1200+80a1τ=8000a1;
60+2a1τ=240a1; |⋅40
1200+80a1τ=8000a1;
2400+80a1τ=9600a1;
−1200=−1600a1;
a1=0.75m/s2;
2a1τ=240a1−3v;
τ=2⋅0.75240⋅0.75−60=80s;
c)a1=0.75m/s2;
Answer: b)τ=80s;
c)0.75m/s2.
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