The driver of a car is going 30 m/s slams on his brakes, the car's tires lock, and the car skids to a stop on a level roadway. The kinetic coefficient of friction between its tires and the pavement is 0.75 as it slides. Draw a free body diagram of the car, find its acceleration (deceleration), and find how far the car slides before stopping.

Question

The driver of a car is going 30 m/s slams on his brakes, the car's tires lock, and the car skids to a stop on a level roadway. The kinetic coefficient of friction between its tires and the pavement is 0.75 as it slides. Draw a free body diagram of the car, find its acceleration (deceleration), and find how far the car slides before stopping.

Answer 1

Assuming uniform deceleration and constant friction (independent of speed).

Equate energies:
(1/2)mu²=μmgX
X=distance
u=30 m/s
g=9.81 m/s²
Solve for X=61.1621 m
Acceleration
v²-u²=2aX
solve for
a=(0²-30²)/(2*61.1621)
=-7.3575 m/s²

Answer 2

To draw a free body diagram of the car, we need to consider the forces acting on it. In this case, there are two main forces:

1. The force of gravity (mg), which acts vertically downward.
2. The force of friction (Ff), which acts in the opposite direction to the motion of the car.

Now, let's find the acceleration (deceleration) of the car:

Using Newton's second law of motion, we can calculate the net force acting on the car. The net force can be expressed as the difference between the force of friction (Ff) and the force of gravity (mg):

Net force (Fnet) = Ff - mg

Since the car is coming to a stop, its final velocity will be zero (v = 0). The initial velocity (u) is given as 30 m/s. Using the equation for acceleration:

Acceleration (a) = (v - u) / t

Since we are assuming the car comes to a stop, the time (t) taken to stop is not known. However, we can use the fact that the car slides to find the relation between acceleration and distance.

The equation linking acceleration (a), initial velocity (u), and distance (s) is:

s = (u^2) / (2a)

Now, let's calculate the acceleration:

Acceleration (a) = (0 - 30 m/s) / t (since final velocity, v = 0)
Acceleration (a) = -30 m/s^2 / t

Now, using the equation for distance, we can find how far the car slides before stopping:

s = (u^2) / (2a)
s = (30 m/s)^2 / (2 * (-30 m/s^2 / t))
s = (900 m^2s^(-2) / (-60 m/s^2 / t)
s = -15 t (simplified using unit analysis)

Since we don't know the time it takes for the car to stop (t), we can't determine the exact distance. However, we can determine the relation between distance and time, which is s = -15t. Therefore, the distance will depend on the time taken to stop.