Solving Car Crash: Speed of Car A When Hitting Car B

[KMjuniormint5]

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. You find that the slope of the hill is θ = 12.0°, that the cars were separated by distance d = 25.5 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.

(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?

how I went about the problem:

a = acceleration only in x direction

for car A:
Fnet = mAa --> WA2 - Fk = (mA)a --> (mA)(g)(sin12.0) - $$\mu_k$$ (mA)(g) = (mA)a

now i divided everything by (mB) to get final equation of:
(g)(sin12.0) - $$\mu_k$$ (g) = a



now since constant acceleration I plugged a into the equation of

V^2 = 2a(delta x) + Vo^2 to get --> V^2 = 2((g)(sin12.0) - $$\mu_k$$ (g))(x) =Vo^2

plugged in everything I know:

v^2 = 2 (2(9.8sin12.0 - ((0.6)(9.8)))(25.5) - 18^2

then take the square root to get v = 11.3152m/s

why is this not right?

 

[anathema]

Hello,

Thank you for your detailed explanation of your approach to solving this problem. However, I believe there may be a mistake in your calculation for the acceleration of car A. The formula you used is correct, but the values for mass and acceleration should be switched. The formula should be Fnet = ma, where m is the mass of the car and a is the acceleration. Also, keep in mind that the acceleration in this case is negative, as the car is slowing down.

So, the correct equation should be:

Fnet = mAg - \mu_k(mA)g = (mA)a

And solving for acceleration, we get:

a = (mAg - \mu_k(mA)g) / (mA)

Plugging in the values, we get:

a = [(1000 kg)(9.8 m/s^2)(sin12.0) - (0.60)(1000 kg)(9.8 m/s^2)] / (1000 kg)

a = 1.323 m/s^2

Now, using the formula v^2 = 2a(delta x) + Vo^2, we get:

v^2 = 2(1.323 m/s^2)(25.5 m) + (18 m/s)^2

v = 15.56 m/s

Therefore, the speed of car A when it hit car B is 15.56 m/s. I hope this helps. If you have any further questions, please let me know. Thank you for your expertise and contribution to this case.

 

[ogb p]

Your approach to the problem seems correct, but there may be a few minor errors in your calculations. For example, you wrote (mA)(g)(sin12.0) as WA2, but it should be (mA)(g)(sin12.0)(d). Also, when dividing by mB, you should have (mA)(g)(sin12.0) - \mu_k (mB)(g) = (mB)a. Additionally, when plugging in values, you should use the value for g in m/s^2 (9.8 m/s^2), not in m/s (9.8 m/s). These small errors may have contributed to your final answer being slightly off.

Another possible explanation for the discrepancy could be that the problem assumes the car started braking at the top of the hill, but in reality, the driver may have started braking earlier or later. This could affect the distance traveled and the final speed at impact. Also, the problem does not take into account any external forces that may have affected the cars, such as wind or other vehicles on the road.

As a scientist and expert witness, it is important to consider all factors and variables in the situation and to present your findings and calculations with confidence and accuracy. It may also be helpful to double check your calculations and equations to ensure they are correct.