1) [tex]1.028 m/s^2[/tex]
We can solve this part by using Newton's second law:
[tex]F=ma[/tex] (1)
where
F is the net force
m is the mass
a is the acceleration
There are two forces acting on the boat:
[tex]F_1= 2.90 \cdot 10^3 N[/tex] forward
[tex]F_2 = 1.69\cdot 10^3 N[/tex] backward
So the net force is
[tex]F=2.90\cdot 10^3-1.69\cdot 10^3=1.21\cdot 10^3 N[/tex]
We know that the mass of the boat is
m = 1177.5 kg
So we can now use eq.(1) to find the acceleration:
[tex]a=\frac{F}{m}=\frac{1.21\cdot 10^3}{1177.5}=1.028 m/s^2[/tex]
2) 161.0 m
We can solve this part by using the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
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)
[tex]a=1.028 m/s^2[/tex]
Substituting t = 17.7 s, we find the distance covered:
[tex]s=0+\frac{1}{2}(1.028)(17.7)^2=161.0 m[/tex]
3) 18.2 m/s
The speed of the boat can be found with the following suvat equation
[tex]v=u+at[/tex]
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)
[tex]a=1.028 m/s^2[/tex]
And substituting t = 17.7 s, we find the final velocity:
[tex]v=0+(1.028)(17.7)=18.2 m/s[/tex]
And the speed is just the magnitude of the velocity, so 18.2 m/s.
