V_final = v_initial + a*t
V_final = 15 + 3 * 4
V_final = 27 m/s
Applying the definition of acceleration, as the rate of change of velocity, we can extract the value of the velocity v at a given time t, as follows:
v = v₀ + a t (1)
As we already know v₀ and t, our single unknown is the acceleration a.
Once the friver slammed on the brakes, there exists an external force, due to friction, which is the only external force acting in the horizontal direction, producing a deceleration.
This friction force, is equal to the product of the coeficient of kinetic friction, times the normal force.
In this case, the normal force is numerically equal to the gravity force, which we call weight.
This force opposes to the direction of the initial velocity, as it always opposes to the relative movement between both surfaces in contact.
Applying Newton's 2nd Law, we get:
Fnet = ma ⇒ Ff = m a ⇒ -μk . m . g = m.a
Solving for a :
a = -μk . g = -0.68 . 9.8 m/s² = -6.66 m/s²
Replacing in (1)
v = 15.4 m/s +(-6.66) m/s². 1.11 s = 8.07 m/s
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
= Coefficient of kinetic friction = 0.63
m = Mass
f = Frictional force
Equation of motion
The speed of the automobile after the time has elapsed is 10.774625 m
coefficient of kinetic Friction
Let final velocity be
The last one is the correct answer to this question that you need me to answer for you
The speed of the automobile after 1.43s is 10