Let the times of motion during three different parts of the distance be t1,t2,t3, the constant speed after the acceleration v1, and accelerations during first and third intervals be a1,a3. Then the corresponding displacements and speeds during three parts of the way are:
1) s1=2a1t12, v1=a1t1
2) s2=v1t2
3) s3=v1t3−2a3t32,0=v1−a3t3.
Total distance is s=s1+s2+s3=2a1t12+v1(t2+t3)−2a3t32.
From 1) and 3), v1=a1t1=a3t3, from where a3=t3t1a1. Substituting the last expression for a3 and v1=a1t1 from 1) into s, obtain:
s=2a1t12+a1t1(t2+t3)−2a1t1t3, from where a1=2t12+t1t2+2t1t3s=0.1min2km.
The steady speed is hence v1=a1t1=0.1min2km⋅10min=1minkm.
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