You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which
was stopped at a red light along a road headed down a hill (Fig. 1). You find that the slope of the hill is Ɵ = 120
, that
the cars were separated by distance d = 24.0 m when the driver of car A put the car into a slide (it lacked any
automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.0 m/s. With what
speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60 (dry road surface) and (b) 0.10 (road
surface covered with wet leaves)?
To find the speed at which car A hits car B in both cases, we need to calculate the deceleration of car A due to friction and then use the equations of motion.
First, let's calculate the deceleration (acceleration due to friction) of car A. The force of friction can be calculated using the equation:
F_friction = μN
where μ is the coefficient of kinetic friction and N is the normal force.
The normal force (N) can be calculated using the equation:
N = mgcos(Ɵ)
where m is the mass of car A and g is the acceleration due to gravity (approximated as 9.8 m/s^2).
Now, the force of friction can be calculated as:
F_friction = μmgcos(Ɵ)
Next, we can use Newton's second law of motion to relate the force of friction to the acceleration:
F_friction = ma
Combining the above equations, we get:
μmgcos(Ɵ) = ma
We can solve this equation for acceleration (a):
a = μgcos(Ɵ)
With the acceleration known, we can now use the equations of motion to find the final velocity of car A.
For a car starting from an initial velocity (v0) and coming to a stop, the equation relating the final velocity (v), initial velocity (v0), acceleration (a), and distance (d) is given by:
v^2 = v0^2 + 2ad
We want to find the final velocity (v) when the distance (d) is equal to the initial separation distance between the two cars (24.0 m).
Plugging in the values we have:
v^2 = v0^2 + 2ad
For part (a), where the coefficient of kinetic friction is 0.6:
a = μgcos(Ɵ)
v^2 = v0^2 + 2(μgcos(Ɵ))d
Substituting the given values:
v^2 = (18.0 m/s)^2 + 2(0.6)(9.8 m/s^2)cos(120°)(24.0 m)
v^2 = 324 m^2/s^2 + 2(0.6)(9.8 m/s^2)(-0.5)(24.0 m)
Now, solve for the final velocity (v). Take the square root of both sides of the equation:
v = √(324 m^2/s^2 + 2(0.6)(9.8 m/s^2)(-0.5)(24.0 m))
Perform the calculation to get the final velocity.
For part (b), where the coefficient of kinetic friction is 0.10:
Repeat the above steps, using the given coefficient of kinetic friction and solving for the final velocity.
Now, you have the final velocities for both cases (a) and (b) when car A hits car B.