   Chapter 4, Problem 11P

Chapter
Section
Textbook Problem

A boat moves through the water with two forces acting on it. One is a 2.00 × 103-N forward push by the water on the propeller, and the other is a 1.80 × 103-N resistive force due to the water around the bow. (a) What is the acceleration of the 1.00 × 103-kg boat? (b) If it starts from rest, how far will the boat move in 10.0 s? (c) What will its velocity be at the end of that time?

(a)

To determine
The acceleration of the boat.

Explanation

Given info: The magnitude of two forces are 2.00×103N and 1.80×103N . The mass of the boat is 1.00×103kg . The initial velocity of the fish is 3.00 m/s and the final velocity is 6.00 m/s. Time taken is 10 s.

The formula to calculate the acceleration is,

a=Fm (I)

• F is the vector sum of all the forces acting.
• m is the mass of the boat.

The vector sum of all the forces is,

F=(2.00×103N)(1.80×103N)=2×102N

The vector sum is the difference between the two forces because the directions of the forces are opposite

(b)

To determine
The distance travelled by the boat.

(c)

To determine
The velocity of the boat at the end of time.

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