- #1
theunloved
- 43
- 1
A 1000kg boat is traveling at 90 km/h when its engine is shut off. The magnitude of the frictional force fk between boat and water is proportional to the speed V of the boat: fk = 70V where V is in meters per second and fk is in Newtons. Find the time required for the boat to slow to 45 km/h.
Attempted to solve:
Take upwards as positive for y direction, x + from left to right
From Newton's law 2, we have:
-fk = ma
a = -fk/m = -70Vi / m
(Vi = V initial)
V = Vi + at
---------> t = (V - Vi) /a
t = - m(V - Vi) / 70Vi (h)
with m = 1000kg, V = 45km/h, Vi = 90 km/h
However, my answer is not correct from the book. Can someone point out where I did wrong ? Thanks
Attempted to solve:
Take upwards as positive for y direction, x + from left to right
From Newton's law 2, we have:
-fk = ma
a = -fk/m = -70Vi / m
(Vi = V initial)
V = Vi + at
---------> t = (V - Vi) /a
t = - m(V - Vi) / 70Vi (h)
with m = 1000kg, V = 45km/h, Vi = 90 km/h
However, my answer is not correct from the book. Can someone point out where I did wrong ? Thanks