Respuesta :
Answer:
378 m
Explanation:
First of all, we have to calculate Sam's acceleration. By using Newton's second law, we can write that the resultant of the forces acting on Sam is equal to the product between his mass and his acceleration (a):
[tex]F-\mu mg = ma[/tex]
where:
F = 170 N is the thrust, pushing Sam forward
[tex]\mu mg[/tex] is the force of friction, acting backward against Sam, with
[tex]\mu = 0.1[/tex] being the coefficient of friction
m = 71 kg is Sam's mass
g = 9.8 m/s^2 is the gravitational acceleration
Solving the equation for a, we find the acceleration:
[tex]a=\frac{F-\mu mg}{m}=\frac{170 N-(0.1)(71 kg)(9.8 m/s^2)}{71 kg}=1.4 m/s^2[/tex]
So, for the first 15 seconds, until he runs out of fuel, he accelerates with this acceleration. Therefore, the distance covered in this part of the trip is
[tex]S_1 = \frac{1}{2}at^2=\frac{1}{2}(1.4 m/s^2)(15 s)^2=157.5 m[/tex]
And the speed reached by Sam after these 15 seconds is
[tex]u=at=(1.4 m/s^2)(15 s)=21 m/s[/tex]
Then, the skis run out of fuel; so now there is no more thrust force, and Newton's second law simply becomes
[tex]\mu mg = ma[/tex]
So, the new acceleration is
[tex]a=-\mu g=-(0.1)(9.8 m/s^2)=-1.0 m/s^2[/tex]
So the distance covered by Sam in this second part of the trip until he stops is given by
[tex]v^2-u^2=2aS_2[/tex]
where
v=0 is the final speed
u = 21 m/s is Sam's initial speed
a = -1.0 m/s^2 is the acceleration
Solving for S2, we find
[tex]S_2=\frac{v^2-u^2}{2a}=\frac{0-(21 m/s)^2}{2(-1.0 m/s^2)}=220.5 m[/tex]
So, the total distance travelled by Sam is
[tex]S=157.5 m+220.5m=378 m[/tex]
The distance covered by Sam within the first 15 seconds is 158.625 m.
seconds and after 15 seconds is 228.2 m.
The total distance covered by Sam is 386.825 m.
What is the distance?
The distance can be defined as the total physical path covered by an object.
Given that the mass m of Sam is 71 kg and the force F is 170 N. The coefficient of friction is 0.1.
The net force is given below.
[tex]F_N = F - F_f[/tex]
Where FN is the net force and Ff is the friction force.
[tex]F_N = 170 - \mu mg[/tex]
[tex]F_N = 170 - 0.1\times 71\times 9.8[/tex]
[tex]F_N = 100.42\;\rm N[/tex]
The acceleration of Sam is calculated as given below.
[tex]F_N = ma[/tex]
[tex]100.42 = 71\times a[/tex]
[tex]a = 1.41\;\rm m/s^2[/tex]
- For the first 15 seconds.
The distance covered by Sam with the same acceleration is given below.
[tex]S = ut + \dfrac {1}{2}at^2[/tex]
Where u is the initial velocity which will be zero and t is the time interval.
Then,
[tex]S = \dfrac {1}{2}at^2[/tex]
[tex]S = \dfrac {1}{2}\times 1.41\times 15^2[/tex]
[tex]S = 158.625\;\rm m[/tex]
The distance covered by Sam within the first 15 seconds is 158.625 m.
seconds.
- After 15 Seconds
The velocity after 15 seconds is given as,
[tex]u' = at[/tex]
[tex]u' = 1.41\times 15[/tex]
[tex]u'=21.15\;\rm m/s[/tex]
The skis run out of fuel after 15 seconds, which means that there is no more thrust force. Then the acceleration will be given as,
[tex]-\mu mg = ma'[/tex]
[tex]-0.1\times 71\times 9.8 = 71\times a'[/tex]
[tex]a' = -0.98\;\rm m/s^2[/tex]
So the distance covered by Sam after 15 seconds until he stops is given as,
[tex]v'^2 -u'^2 = 2a'S'[/tex]
Where v' is the final velocity when he stops so v'=0.
[tex]0^2 - 21.15^2 = 2\times(- 0.98)\times S'[/tex]
[tex]S'= 228.2\;\rm m[/tex]
The distance covered by Sam after 15 seconds is 228.2 m.
The total distance covered is given as,
Distance = S + S'
Distance = 158.625 + 228.2
Distance = 386.825
Hence we can conclude that the total distance covered by Sam is 386.825 m.
To know more about the distance and acceleration, follow the link given below.
https://brainly.com/question/14363745.
