m/s2 for 10.0 s, then slows down to a stop with a constar SeA len sioning...

Question

Question

Answer 1

Answer #1

Distance traveled while acceleration

= 1/2at² =1/2*2.20*10*10=110m

Distance travelled while stopping= s=v²/2a

Where v is 22 m/s

S= 22*22/2*3.05 =79.34

Total distance=110+79.34=189.34

Answer 2

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