Question

(Part 1 of 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:

$a=\frac{F}{m}=\frac{1.21\cdot 10^3}{1177.5}=1.028 m/s^2$

2) 161.0 m

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

$s=ut+\frac{1}{2}at^2$

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:

$s=0+\frac{1}{2}(1.028)(17.7)^2=161.0 m$

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.