Was the Driver Speeding Before the Crash?

[Logan Batson]

Homework Statement

"Based on the following data, determine if the driver who crashed was driving over the speed limit"

The relevant data given is:

-A 2002 Volvo t-bones a truck in an alley with a speed limit of 20km/h

-An eyewitness heard tires screech, then a loud bang, and also said the driver hit the brakes at the last second, and was going too fast to stop

-Skid marks created by the car were 18.3m in length, with a consistent acceleration of 3.9m/s^2

-A doctor states in a medical record that the bruising and lacerations from the crash are from an impact of 25g's

-The doctor also said that the crash couldn't have been longer than 1/20th of a second

Homework Equations

v = d/t

v_f = v_i + at

d = ( (v_f - v_i) / 2)*t

d = ( v_i*t ) + ( .5*a*t² )

d = ( v_f*t ) - ( .5*a*t² )

v_f² = v_i² + 2*a*d

The Attempt at a Solution

[/B]
My thought process so far was to work backwards with a d/t, v/t, and a/t graph. I believe there is four "sections" to the graphs, before the crash, breaking, the crash, and the rolling back from the impact.

I believe the skid marks relate to the breaking motion, while the acceleration given is related to the reverse motion after the crash. I believe the crash portion of the graphs is all zero, except displacement.

I'm stuck for how to continue after this point. The before and after the crash sections will have an unknown time, and seem impossible to solve without the time.

 

[Mohammeddev]

consider this we knew that :
Vf = 0m/s
a = 3.9m/s²
d = 18.3m

so we need to find equation that not include the time to find the initial velocity :
v² = Vi² + 2ad

i hope i understand your question right

 

[Logan Batson]

consider this we knew that :
Vf = 0m/s
a = 3.9m/s²
d = 18.3m

so we need to find equation that not include the time to find the initial velocity :
v² = Vi² + 2ad

i hope i understand your question right

I believe that's correct yes, however I would not know if this initial velocity would be the start of the breaking section, or the start of the reverse section

 

[Mohammeddev]

we knew it's the start of the breaking section because the displacement we calculate which is : 18.3m is the start of the skid which is the start of the breaking

 

[Orodruin]

consider this we knew that :
Vf = 0m/s
a = 3.9m/s2
d = 18.3m

so we need to find equation that not include the time to find the initial velocity :
v2 = Vi2 + 2ad

i hope i understand your question right

If this were true there would have been no accident. The car would have stopped just in time to just touch the truck. You need to use more information.

 

[DrClaude]

This doesn't sound like a textbook problem, but rather as an accident reconstruction. We don't allow that at PF, not just for legal reasons, but also because there is never enough data.

Thread closed.

Edit: After some discussion, there is a feeling that this is a legitimate problem, so I'll reopen the thread. Sorry for mess.

 

[jbriggs444]

so we need to find equation that not include the time to find the initial velocity :
$v^2 = V_i^2 + 2ad$

If you are trying to compute the velocity, $v$, at the start of the braking section in terms of impact velocity, $V_i$, then I would agree with that equation.

I believe the skid marks relate to the breaking motion, while the acceleration given is related to the reverse motion after the crash.

I read the problem differently. The acceleration, a, is the acceleration due to braking.

Skid marks created by the car were 18.3m in length, with a consistent acceleration of 3.9m/s^2

Reasoning:

1. Since we are not given a displacement of the mangled resulting car and truck remains from the end of the skid marks, the acceleration between the time of impact and the time when the remnants come to rest, knowing the acceleration during that phase would do us no good.

2. Since we do know the displacement of the car during the braking phase, the acceleration during that phase does us some good and is crucial to determine the velocity prior to the braking phase.

3. The deceleration due to skidding tires on pavement tends to be predictable. The deceleration of mangled piles of debris tends not to be.

4. Homework problems normally include all information required to solve them and usually do not include excess information.

5. The information about skid mark length and deceleration was given in the same bullet point.

@Orodruin seems to believe that we have been given information about the impact velocity. I do not see anything to support that in the problem statement.

The doctor says that the duration of the impact could have been between 0 and 0.05 seconds. I suspect that this is a mis-statement by the person who set the problem and that the intent was that the impact had to be 0.05 seconds or greater. However, we solve problems as they are set, not as they are meant to be set. [Or we could solve both ways].

-The doctor also said that the crash couldn't have been longer than 1/20th of a second

The eyewitness says that the initial velocity was high enough that a crash was inevitable. But makes no claim about the resulting impact velocity.

-An eyewitness heard tires screech, then a loud bang, and also said the driver hit the brakes at the last second, and was going too fast to stop

 

[Orodruin]

I do not see anything to support that in the problem statement.

I believe the problem is intended to be solved using the information about the crash acceleration and time to deduce the impact velocity (or with the current information, bound it from above and below).

 

[scottdave]

Remember that you don't have to calculate the actual speed, just find the minimum speed that would cause these events.. if that is higher than 20 k'm/h then you know the drill er had to be speeding

 

[Orodruin]

Remember that you don't have to calculate the actual speed, just find the minimum speed that would cause these events.. if that is higher than 20 k'm/h then you know the drill er had to be speeding

Based on the actual numerical values in the problem, I would suggest that there is an additional (intentional) pitfall.