A boat moves through the water with two forces acting on it. One is a 2.62x10³ Newtons (N) forward push by the motor, and the other is a 1.70 x 10³ N resistive force due to the water.
a) What is the acceleration of the1349.4 kg boat? Answer in units of m/s²
b) If it starts from rest, how far will it move in 13.5 s? Answer in units of m.
c) What will its speed be at the end of this time interval? Answer in units of m/s.
My teacher is not very understanding. I have been absent for a week for a family funeral and I really do not understand what has been going on in physics class. Please help me solve any of these or all of them and show how you got the answer. ANYTHING helps. Thank you so much!
2007-01-30 14:59:27 · 2 answers · asked by susie q 2 in Science & Mathematics ➔ Physics
It's been a long time since I have done this, but you need 2 things: f = ma, and the total acceleration. The forward force is 2.62x10³, the backward force (in exactly the opposite direction) is 1.70 x 10³, so the net force is 0.92 x 10³ in the forward direction.
So, f = ma, you know m = 1349.4 kg, f = 0.92 x 10³, so a = f/m, a = .682 (I think the units work out, but I forget exactly what newtons are. Let's assume this is in m/s^2 - I think it is, but you should check to make sure)
c) Starting from 0, with constant acceleration of .682 m/s^2, after 13.5 seconds its speed will be 13.5*.682 = 9.207 m/s
b) Distance covered while inder constant acceleration is average speed*time, average speed is (0 + 9.207)/2, so distance is 62.14 meters.
Good luck learning this, and condolances for your family's loss.
*** check my work for errors ***
2007-01-30 15:22:31 · answer #1 · answered by sofarsogood 5 · 2⤊ 0⤋
a) Fnet=ma so (2620-1700)/1349.4=a
a=.682m/s^2
b) s=.5at^2+vit so s=.5(.682)(13.5)^2+0
s=62.147m
c)vf=vi+at so vf=0+(.682)(13.5)
vf=9.207
2007-01-30 23:14:45 · answer #2 · answered by climberguy12 7 · 2⤊ 0⤋