Respuesta :
Answer:
The shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.4 m/s = 55.125 m = 55.13 m
Explanation:
Coefficient of kinetic friction = 0.8
Frictional force can be evaluated using μN
where N is the normal reaction = mg
Fr = 0.8 m × 9.8 = 7.84 m
The car stops with th help of friction, this friction manifests in a form of decelerating force.
This frictional force will be equal to the ma force,
ma = 7.84 m
a = 7.84 m/s²
But Since it's deceleration, this acceleration is - 7.84 m/s²
So, using the equations of motion,
u = initial velocity = 29.4 m/s
v = final velocity = 0 m/s (since it comes to a stop finally)
x = distance travelled before stopping
a = - 7.84 m/s²
v² = u² + 2ax
0² = 29.4² + 2(-7.84)(x)
15.68 x = 864.36
x = 55.125 m = 55.13 m
Answer:
54.02m
Explanation:
According to Newton's second law of motion;
∑F = m x a ------------------(i)
Where;
∑F = effective force acting on a body
m = mass of the body
a = acceleration of the body
(i) According to the question, the only force acting on the horizontal (direction of motion) is the frictional force ([tex]F_{R}[/tex]) between the tires of the automobile and the dry pavement and it is directed opposite the direction of motion. Therefore, it is negative.
i.e ∑F = - [tex]F_{R}[/tex]
Where;
[tex]F_{R}[/tex] = μ[tex]_{K}[/tex] x N
μ[tex]_{K}[/tex] = coefficient of kinetic friction.
N = normal reaction = m x g
m = mass of the automobile as depicted by the tires and
g = acceleration due to gravity = 10m/s²
=> ∑F = - μ[tex]_{K}[/tex] x m x g
Substitute the value of ∑F into equation (i) as follows;
- μ[tex]_{K}[/tex] x m x g = m x a
Divide through by m;
- μ[tex]_{K}[/tex] x g = a ------------------(ii)
Given;
μ[tex]_{K}[/tex] = Coefficient of kinetic friction between tires and dry = 0.800
Substitute this value into equation (ii) as follows;
- 0.800 x 10 = a
a = - 8m/s²
Therefore, the acceleration of the automobile is -8m/s² [The negative sign shows deceleration]
(ii) Now, the distance (s) is gotten from one of the equations of motion as follows;
v² = u² + 2as ---------------------(iii)
Where
v = final velocity of the automobile = 0 [since the automobile is stopped]
u = initial velocity of the automobile = 29.4m/s
a = acceleration of the automobile = - 8m/s²
Substitute these values into equation (iii) as follows;
0² = 29.4² + [2(-8) x s]
0 = 864.36 - 16s
16s = 864.36
Solve for s;
s = [tex]\frac{864.36}{16}[/tex]
s = 54.02m
Therefore, the shortest distance in which the automobile can be stopped by locking the brakes when traveling at 29.4m/s is 54.02m
