A boat going against a current of 6km/hr?
August 15, 2008 – 10:00 pmThere ia a underway of 6 km/hr. It takes a dish threesome hours to go digit artefact and digit hours to become back. It travels direct into the underway on the prototypal activate and with the underway on the ordinal trip.
would the boats pace be in ease water. I got 30 km/hr using digit equations with digit unknowns and change what do you get
One Response to “A boat going against a current of 6km/hr?”
sorry: corrected from first post silly! Thank you steve.
distance up distance back distance net speed x time d=(v-6)*3 distance up d=(v+6)*2 distance back solving for v: v=30 km/hr
By Al P on Aug 15, 2008