Car Crash Problem: Was Driver Speeding?

[calcite]

Homework Statement

There are two cars and the situation is located at a crossroad. car 1 is traveling south and car 2 is traveling east. For some strange reason car 2 has slammed the brakes and stopped right in the middle of the intersection, causing car 1 to 't-bone' crash into car 2. the mass of the cars are given:

mass(car1) = 1875 kg
mass(car2) = 1051 kg

the distance of car1's skid marks are = 16m

The speed limit on the crossroad is 50 km/h, the question of this problem is, was the driver of car1 speeding?

Homework Equations

s = ut + 1/2at^2
F(friction) = \mumg
f=ma
P=mv ?

The Attempt at a Solution

i think what the question is asking for is 'u', the speed car 1 was doing before he applied the brakes.
Therefore i used s = ut + 1/2at^2
however i am not given acceleration or time, 's' is the displacement of when car1 applied the brakes to when he hit car2, therefore 's' is the distance of the skidmarks ( assuming the car make skidmarks as soon as the driver applied the brakes :P) s = 16m

Using the Friction equation, F(friction) = \mumg i found the acceleration

F = \mumg
ma = \mumg
a = \mug
a = 0.8 x 9.8
= 7.84 m s-2

so all i need to do now is find the time of the braking , but i have tried everything and i cannot find anyway of doing it...i feel like I am at a dead end, what should i do?

 

[tiny-tim]

Welcome to PF!

… car 1 is traveling south and car 2 is traveling east. For some strange reason car 2 has slammed the brakes and stopped right in the middle of the intersection, causing car 1 to 't-bone' crash into car 2. the mass of the cars are given:

mass(car1) = 1875 kg
mass(car2) = 1051 kg

the distance of car1's skid marks are = 16m
…
i think what the question is asking for is 'u', the speed car 1 was doing before he applied the brakes.
Therefore i used s = ut + 1/2at^2

Hi calcite! Welcome to PF!

Hint: use the ordinary collision equations, and then use work-energy (force x distance = energy lost).

 

[crumpet]

Go to QUT much?

 

[calcite]

are you saying to use conservation of momentum?

its inelastic so...

mass(1) x velocity (1) + mass(2) x velocity(2) = (mass(1) + mass(2)) velocity (f)

i don't understand what we use for the velocities...

 

[calcite]

Go to QUT much?

lol yes, who are you?

 

[tiny-tim]

are you saying to use conservation of momentum?

its inelastic so...

mass(1) x velocity (1) + mass(2) x velocity(2) = (mass(1) + mass(2)) velocity (f)

i don't understand what we use for the velocities...

Hi calcite!

(I'm assuming car 1 and car 2 have the same velocities after the collision:)

You know the friction force and the distance, so you can work out the energy of the cars, and from that you get the velocities v1f = v2f immediately after the collision.

And you know v2i = 0.

So your mometum equation gives you v1i.

 

[crumpet]

lol yes, who are you?

Haha, it's a secret :)
I'm here trying to find a means for one of those quiz questions, but I couldn't help but click on your "Car Crash" topic.

 

[Simo43]

Can you be of any help, I need a simple (if that is possible) equation to help me determine the weight increase of a 12 stone occupant of a car doing 30mph coming to a dead stop?