Asked by dy
A passenger train takes 3 hours less for a journey of 360km . if its speed increased by 10km/hr from its usual speed . what is its usual speed?
Answers
Answered by
Damon
360 = v t
360 = (v+10)(t-3) = vt -3v + 10t -30
so
vt = vt -3v + 10 t - 30
10 t = 3 (v+10)
but t = 360/v
3600/v = 3 v + 30
3 v^2 +30 v - 3600 = 0
v^2 + 10 v - 1200 = 0
v+40)(v-30) = 0
v = 30
360 = (v+10)(t-3) = vt -3v + 10t -30
so
vt = vt -3v + 10 t - 30
10 t = 3 (v+10)
but t = 360/v
3600/v = 3 v + 30
3 v^2 +30 v - 3600 = 0
v^2 + 10 v - 1200 = 0
v+40)(v-30) = 0
v = 30
Answered by
Anonymous
360 = v t
360 = (v+10)(t-3) = vt -3v + 10t -30
so
vt = vt -3v + 10 t - 30
10 t = 3 (v+10)
but t = 360/v
3600/v = 3 v + 30
3 v^2 +30 v - 3600 = 0
v^2 + 10 v - 1200 = 0
v+40)(v-30) = 0
v = 30
360 = (v+10)(t-3) = vt -3v + 10t -30
so
vt = vt -3v + 10 t - 30
10 t = 3 (v+10)
but t = 360/v
3600/v = 3 v + 30
3 v^2 +30 v - 3600 = 0
v^2 + 10 v - 1200 = 0
v+40)(v-30) = 0
v = 30
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