Asked by lee
A locomotive of mass 20000kg is traveling at a speed of 54km/hr on a straight track.if a breaking force of 250N per turn is applied .how far will the locomotive travel before being brought to rest?
My solution:v=0 u=15m/s S=? F=250 M=20000
F=ma
250=20000×a
a=250/20000 =0.0125
V(square)= u(square) +2as
O(square)=15×15 + (2×0.0125)s
0=225+0.025s
-225=0.025s
S=-225/0.025
S=-9000m
My solution:v=0 u=15m/s S=? F=250 M=20000
F=ma
250=20000×a
a=250/20000 =0.0125
V(square)= u(square) +2as
O(square)=15×15 + (2×0.0125)s
0=225+0.025s
-225=0.025s
S=-225/0.025
S=-9000m
Answers
Answered by
bobpursley
Looks good to me, I didn't check math.
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