# How can i get an initial speed without acceleration or time?

How can i get an initial speed without acceleration or time?!?!

## Homework Statement

Hi..heres the problem:
A 4WD travelling East has collided with a stationary car, the 4WD's skid marks are 15m long until the point of collision. The 4WD is sent 35degs South for 15.8m and the car is sent 55degs North for 15m. The car has a mass of 1150kg and the 4WD's is 1600kg. .The coefficient of friction is 0.6.
Find the initial speed of the 4WD before braking in km/h.
this is ALL the info included...

## Homework Equations

s=displacement, mu=0.6, v=final velocity, u=initial velocity, P=momentum, Pi=initial momentum (also just the momentum of 4WD before crash), Pf=final momentum, P'1=momentum of 4WD after collison, P'2=momentum of car after collision, g=gravity(9.8plz), a=acceleration t=time ( numbers after pronumerals signify which vehicle: 4WD=1 , Car=2...dashes between pronumeral and vehicle number mean after the crash eg: P'1)

u= sqr root ( 2mu*g*s)

P=mv

pythagorus= to add the final momentums to get the initial momentum

v^2=u^2 + 2as

s=ut+1/2 at^2 ....(1/2 = half)

v=u+at

## The Attempt at a Solution

to find initial speeds after the crash:
u'1= sqr root:(2*0.6*9.8*15.8)
u'1= 13.63m/s

u'2= sqr root:(2*0.6*9.8*15*)
u'2= 13.28m/2

to find the initial momentum (initial momentum of 4WD)
Pi^2 = (m'1*u'1)^2 + (m'2*u'2)^2
Pi^2=(1600*13.63)^2 + (1150*13.28)^2
Pi= 26623.727 kgm/s
then to find v1= (26623.727)/(1600)
v1= 16.63m/s

from there on i dont know what to do.. I've been advised that the last answer (16.63m/s) was the final speed before the collison...is that right? if so, how can i find the initial speed before braking? im guessing i have to find acceleration, but i dont know how to without time, and visa versa... apparently the initial speed is in the 70's km/h...:S plz help

NascentOxygen
Staff Emeritus

Perhaps you are to assume the coefficient of friction during braking is also 0.6? What answer do you get if you assume that?

thanks!
Perhaps you are to assume the coefficient of friction during braking is also 0.6? What answer do you get if you assume that?
do you mean by using the initial speed formula?

u= √2mugs

u=√(2*0.6*9.8*15)

u= 13.28ms-1

its the wrong answer if thats what you meant because 15m isnt the whole distance the 4WD would have taken to stop as it crashed before it could....but do you think that if i could find the extra amount of distance it would have travelled if it didnt crash into the car, then added that to the braking distance of 15m and then used the u= √2mugs rule that it might work?

If i add the vectors of 15m and 15.8 m i get :
unknown distance after crash if crash didnt happen= √(15^2 + 15.8^2)
= 21.79m
then if i add that to the initial braking distance:
= 21.79 + 15
= 36.79m
and use that in u= √(2mugs)
u= √(2*0.6*9.8*36.79)
u= 20.8ms-1
u= 74.9kmh-1
the speed is said to be in the 70's...so maybe this is correct?!? =D

NascentOxygen
Staff Emeritus

Let's assume your figure of v1= 16.63m/s is correct (because I haven't checked it).

Before the collision, can we say the vehicle had decellerated over a distance of 15m by a braking force of μ·m·g?

Use v² = u² + 2as

You won't obtain 70kph, but I'm yet to be convinced of the correct answer. I know this question will be solved using law of conservation of momentum and/or some other higher principles which i haven't read yet..and i was thus trying with basic kinematics principles

The car that is sent north travels 15 meters and taking final velocity to be zero and a=-6ms-2 we can get the initial velocity=6√5 ms-1.. Now my doubt is-(for the same car) v know what velocity it attained,v know that it attained it from rest..we also know that the time period it took to do so was infinitely small and so can v not somehow figure out the total acceleration it underwent(using calculus maybe-that deals with very small intervals of time..i don't know) and thus the force with which it was hit and thus the velocity with which the body hitting it was travelling with??

Sorry this all may be making no sense but i had to ask!

NascentOxygen
Staff Emeritus

The car that is sent north travels 15 meters and taking final velocity to be zero and a=-6ms-2
Where did a=-6ms⁻² come from?
the time period it took to do so was infinitely small
Not infinitely small, but we have no idea how small.

NascentOxygen
Staff Emeritus

Let's assume your figure of v1= 16.63m/s is correct (because I haven't checked it).
I have checked it now. 16.63m/sec is close enough. You might have to rethink your method if the paths were not exactly perpendicularhttps://www.physicsforums.com/images/icons/icon4.gif [Broken]

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Where did a=-6ms⁻² come from?
.
(0.6 times normal reaction)/mass..i assumed it's coefficient f kinetic friction

but we have no idea how small.
Well that is what u call infinitely small don't you! when you can only imagine how very very small it was.

NascentOxygen
Staff Emeritus