Question

(Part 1 of 3)

A boat moves through the water with two forces acting on it. One is a 2.90×103 N forward push by the motor, and the other is a 1.69×103 N resistive force due to the water.
What is the acceleration of the 1177.5 kg boat?
ANSWER: 1.0276
(Part 2 of 3)
If it starts from rest, how far will it move in 17.7 s?
ANSWER: 160.969
(part 3 of 3)
What will be its speed at the end of this time?
ANSWER: unknown

I NEED ANSWER FOR PART 3

Answer 1

1)
`1.028 m/s^2`

We can solve this part by using Newton's second law:

`F=ma` (1)

where

F is the net force

m is the mass

a is the acceleration

There are two forces acting on the boat:

`F_1= 2.90 \cdot 10^3 N` forward

`F_2 = 1.69\cdot 10^3 N` backward

So the net force is

`F=2.90\cdot 10^3-1.69\cdot 10^3=1.21\cdot 10^3 N`

We know that the mass of the boat is

m = 1177.5 kg

So we can now use eq.(1) to find the acceleration:

`a=(F)/(m)=(1.21\cdot 10^3)/(1177.5)=1.028 m/s^2`

2) 161.0 m

We can solve this part by using the following suvat equation:

`s=ut+(1)/(2)at^2`

where

s is the distance travelled

u is the initial velocity

t is the time

a is the acceleration

Here we have

u = 0 (the boat starts from rest)

`a=1.028 m/s^2`

Substituting t = 17.7 s, we find the distance covered:

`s=0+(1)/(2)(1.028)(17.7)^2=161.0 m`

3) 18.2 m/s

The speed of the boat can be found with the following suvat equation

`v=u+at`

where

v is the final velocity

u is the initial velocity

t is the time

a is the acceleration

In this case we have

u = 0 (the boat starts from rest)

`a=1.028 m/s^2`

And substituting t = 17.7 s, we find the final velocity:

`v=0+(1.028)(17.7)=18.2 m/s`

And the speed is just the magnitude of the velocity, so 18.2 m/s.