A driver sees a horse on the road and applies the brakes so hard that they lock and the car skids to stop in 24 m .The road is level ,and the coefficient of kinetic friction between tires and the road is 0.7.How fast was the car going when the brakes were applied?
weight = m g
so F = -mu m g =- 0.7 * m * 9.81 =- 6.87 m
so acceleration = - 6.87 m/s^2
v = Vi + a t
0 = Vi - 6.87 t
t = Vi/6.87
average speed = Vi/2
so
24 = (Vi/2)(Vi/6.87)
Vi^2 = 329.8
Vi = 18.2 m/s
M*g = 9.8M = Wt. of vehicle = Normal(Fn).
Fp = Mg*sin A = 9.8M*sin 0 = 0. = Force parallel with surface.
Fk = u*Fn = 0.7 * 9.8M = 6.86M N. = Force of kinetic friction.
Fp - Fk = M*a,
0 - 6.86M = M*a,
a = -6.86 m/s^2.
V^2 = Vo^2 + 2a*d = 0,
Vo^2 - 13.72*24 = 0,
Vo = 18.1 m/s.
Well, it seems like this driver really had to rein in their speed! Let's calculate it in a horsy way. We know that the car skidded to a stop, so it came to rest. To find the initial speed, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s since it came to rest)
u = initial velocity (what we need to find)
a = acceleration (due to friction, in this case)
s = distance covered (24 m)
Now, the force of friction can be calculated as:
friction = coefficient of friction * normal force
Since the road is level, the normal force is equal to the weight of the car (mg). And since the car is not accelerating vertically, the normal force is equal to the gravitational force (mg). Therefore, friction = coefficient of friction * mg.
Using 0.7 as the coefficient of friction, we can plug in the values:
friction = 0.7 * (mass of the car * g)
Now, the force of friction can be related to the acceleration by Newton's second law: force = mass * acceleration. So:
0.7 * (mass of the car * g) = mass of the car * acceleration
The mass of the car cancels out, leaving us with:
0.7 * g = acceleration
Now, let's substitute this acceleration value, along with the distance covered, into the first equation:
0^2 = u^2 + 2 * (0.7 * g) * 24
Since the car was initially at rest, the initial velocity (u) squared cancels out:
0 = 2 * (0.7 * g) * 24
Dividing both sides by 2 * 24 * 0.7, we get:
0 = g
Whoops, it looks like we're stuck in a tricky situation here! According to my calculations, the acceleration due to friction doesn't affect the final result. So, it seems we're left with a bit of a mathematical muddle. Perhaps it's best to consult an expert or try a different approach to solve this problem!
To find the initial speed of the car, we can use the formula for frictional force:
Frictional force = (coefficient of kinetic friction) * (normal force)
The normal force equal to the weight of the car, which can be calculated using:
Weight = mass * gravity
Now, let's break down the problem step-by-step:
Step 1: Determine the normal force.
Since the road is level, the normal force is equal to the weight of the car.
Step 2: Calculate the weight of the car.
You need the mass of the car to determine its weight.
If the mass of the car is given, use that value. Otherwise, assume a reasonable value for the car's mass.
Step 3: Calculate the frictional force.
Frictional force = (coefficient of kinetic friction) * (normal force)
Step 4: Apply Newton's second law.
The net force acting on the car in the horizontal direction is the frictional force.
The net force is equal to the mass of the car multiplied by its deceleration.
Using Newton's second law, calculate the deceleration of the car.
Step 5: Use the kinematic equation.
Since the car started with an initial speed and ended with a final speed of 0 m/s, we can use the equation:
(v^2 - u^2) = 2aS
where:
- v = final velocity (0 m/s)
- u = initial velocity (unknown - what we want to find)
- a = acceleration (deceleration of the car)
- S = distance traveled (24 m)
Solve this equation for the initial velocity (u) of the car.
Let's calculate each step step-by-step.
To find the initial speed of the car before the brakes were applied, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s as the car comes to a stop)
u = initial velocity (what we want to find)
a = acceleration (in this case, deceleration due to braking)
s = distance traveled (24 m)
Rearranging the equation, we get:
u^2 = v^2 - 2as
Substituting the given values:
u^2 = 0^2 - 2 * 0.7 * 24
Simplifying:
u^2 = -33.6
Since velocity cannot be negative, the negative sign indicates the deceleration. We take the square root of 33.6:
u = √33.6
Therefore, the car was going approximately (√33.6) m/s when the brakes were applied.