Physics
posted by Jason .
A car traveling 56 km/h is 24 m from a barrier when the driver slams on the brakes. The car hits the barrier 2s later
a) what is the car's constant deceleration before the impact?
b)How fast is the car traveling at impact?
I don't get the impact parts...
I think you have to use one of these formulas:
v=v0+at
xx0=v0t+1/2at^2
v^2v0^2=2a(xx0)
xx0=(v0+v)/2 *t
xx0=vt1/2 at^2
But I'm not sure if these are correct...
The question unanswered here is does the car hit the barrier? The velocity at impact is not known, therefor the acceleration cannot be determined (as if the brakes did not stop it). Now if one assumes the car stopped just at the barrier, then the final velocity is zero, so you know acceleration...
Vf^2=Vo^2 + 2*accelertion*distance and you can solve for acceleration.
The question b) is really the question, and it is cannot be determined. We assumed Vf was zero in part a. Something is flawed about the construction of this question.
Look at the formula xx0=(v0+v)/2 *t
Let x0=0 and x=24m and and t=2sec so
2*24m/2sec=v0+v or 24m/s=56m/s + v so v=32m/s I'm really not sure if this makes sense here. If so, then
(vv_0)/t=average deceleration so (3256)/2 = 88/2 =44m/s^2
Be sure to check that I use the formula correctly. See if there's an example in your book like this.
If we use this formula xx0=v0t+1/2at^2 then
24m=56m/s *2sec + (1/2)a(2sec)^2 so
2(24m112m)/4sec^2=44m/sec^2=a which agrees with the first result
If we use v=at+v0 with a=44m/sec^2 and t=2sec, v0=56m/s then
v=88m/s+56m/s=32m/s
It seems to agree also, but I'm not sure about the negative velocity.
Hopefully this helps...
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