v0=0v_0=0v0=0 - A train starts from rest
t=10st=10st=10s - the time with acceleration
a=4−0.3ta=4-0.3ta=4−0.3t
aavr=a(t0=0)+a(t10=10)t10−t0=0+110=0.1a_{avr}=\frac{a(t_0=0)+a(t_{10}=10)}{t_{10}-t_0}=\frac{0+1}{10}=0.1aavr=t10−t0a(t0=0)+a(t10=10)=100+1=0.1
v=v0+aavrt22v=0+0.1⋅1022=5v=v_0+\frac{a_{avr}t^2}{2}\\ v=0+\frac{0.1 \cdot 10^2}{2}=5v=v0+2aavrt2v=0+20.1⋅102=5 m/s
Answer: 5 m/s
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