A car without ABS (antilock brake system) was moving at 16.5 m/s when the driver hit the brake to make a sudden stop. The coefficients of static and kinetic friction between the tires and the road are 0.553 and 0.410, respectively.
(a) What was the acceleration of the car during the interval between braking and stopping?
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(b) How far did the car travel before it stopped?
Vf^2=Vi^2 + 2ad
a=f/m=coeffstatic*mg/m=coeffstatic*g
how far, use the first equation. use a (acceleration as negative)
To find the acceleration of the car during the interval between braking and stopping, we can use Newton's second law of motion, which states that the force applied to an object is equal to its mass times its acceleration (F = ma).
(a) To calculate the acceleration, we first need to find the net force acting on the car. The net force can be determined by subtracting the force of kinetic friction (Fk) from the force of static friction (Fs). The force of static friction can be calculated by multiplying the coefficient of static friction (μs) by the weight of the car (mg), where g is the acceleration due to gravity (9.8 m/s^2).
Fs = μs * mg
Then, we subtract the force of kinetic friction from the force of static friction to find the net force:
Fnet = Fs - Fk
Now, we can substitute the calculated net force into Newton's second law to solve for acceleration:
Fnet = ma
(a) Solution:
1. Calculate the force of static friction:
Fs = μs * mg
= 0.553 * (mass of the car * 9.8)
= 0.553 * (mass of the car * 9.8)
2. Calculate the force of kinetic friction:
Fk = μk * mg
= 0.410 * (mass of the car * 9.8)
= 0.410 * (mass of the car * 9.8)
3. Calculate the net force:
Fnet = Fs - Fk
4. Substitute the net force into Newton's second law:
Fnet = ma
=> (Fs - Fk) = ma
=> (0.553 * (mass of the car * 9.8)) - (0.410 * (mass of the car * 9.8)) = ma
Now, this equation can be solved for acceleration (a). It is important to note that you need the mass of the car to find the acceleration.
(b) To find the distance the car traveled before stopping, we can use the equations of motion. The equation we will use is:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car stops)
u = initial velocity (16.5 m/s)
a = acceleration (from part (a))
s = distance traveled
(b) Solution:
1. Rearrange the equation for distance:
s = (v^2 - u^2) / (2a)
2. Substitute the values into the equation:
s = (0^2 - 16.5^2) / (2 * acceleration)
By substituting the calculated acceleration from part (a), you can solve the equation to find the distance traveled before the car stops.