In coming to a stop, a car leaves skid marks 92 m long on the highway...?

Question

Assuming a deceleration of 7.00 m/s estimate the speed of the car just before braking.

I have tried to work this problem but I am not sure I used the correct equation. Thanks!

Answer 1

Well I see one problem. I hope it's just a typo on your part. Acceleration (or deceleration) is in units of distance/time^2. You say "deceleration of 7.00 m/s". Let me assume you meant to say 7 m/s^2 and I'll continue.

There are 4 motion formulas that you probably have been given. One of them is probably V^2 = Vo^2 + 2*a*s - perhaps not in this exact form. This seems to involve everything you know plus what you need. In your problem final velocity will be zero and you're looking for original velocity. You know acceleration (well actually deceleration) acceleration will be -7 m/s^2, and distance, s=92m. Plug this data in and solve for Vo.

Answer 2

At a decelleration of 7 m/s and skid marks 92m long...........mmmmm
It took 13.14 seconds to stop loosing 7 m each second.
That's about the gist of it.....so far.......
In order to need 13.14 seconds @ 7 m/s to stop, the vehicle had to have an initial volocity of 92 m/s - or 5520 m/m - or 331200 m/h -
or 331 Km/h......206 mph??? Sounds like he was speeding to me.

Answer 3

decreasing speed at 7m/s/s wont equal 13 seconds....

After 1 second 7m
after 2 seconds 14m
so by the time its been 3 seconds elapsed its stopped by 21+14+7= 42m think how long 3 seconds REALLY is... in terms of a vehicle slowing down

Answer 4

The trick in doing all those questions, is to record out what you've. (i.e as decrease than) then record out the equations of action and observe which ones you've sufficient prevalent values for. s = seventy 5 m a = 7.5 ms^-2 v = 0 u = ? i ought to apply v^2 = u^2 + 2 a(s)

Answer 5

ask a cop.

they investigate this kind of stuff.