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
x-x0=v0t+1/2at^2
v^2-v0^2=2a(x-x0)
x-x0=(v0+v)/2 *t
x-x0=vt-1/2 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 x-x0=(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
(v-v_0)/t=average deceleration so (-32-56)/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 x-x0=v0t+1/2at^2 then
24m=56m/s *2sec + (1/2)a(2sec)^2 so
2(24m-112m)/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...
The given question involves finding the car's constant deceleration and its speed at the moment of impact. Let's break down the steps to solve this problem:
a) To find the car's constant deceleration before the impact, we can use the formula v^2 - v0^2 = 2a(x - x0), where v is the final velocity, v0 is the initial velocity, a is the acceleration, x is the final position, and x0 is the initial position.
In this case, the car's initial velocity, v0, is 56 km/h. We need to convert this to m/s, so v0 = 56 km/h * (1000 m/1 km) * (1 h/3600 s) = 15.56 m/s.
The final velocity, v, is 0 m/s, because the car hits the barrier and comes to a stop.
The initial position, x0, is 0 m, as it is the starting point of measurement.
The final position, x, is given as 24 m.
Using the formula v^2 - v0^2 = 2a(x - x0), we can substitute the known values:
0^2 - (15.56 m/s)^2 = 2a(24 m - 0 m)
Simplifying the equation:
-15.56^2 = 48a
Solving for a:
a = (-15.56^2) / 48 ≈ -5.02 m/s^2
Therefore, the car's constant deceleration before the impact is approximately -5.02 m/s^2.
b) To find the speed of the car at the moment of impact, we can use the formula v = v0 + at.
In this case, the initial velocity, v0, is 15.56 m/s. We can use the same value of acceleration, a = -5.02 m/s^2, from part a.
The time, t, at the moment of impact is given as 2 seconds.
Using the formula v = v0 + at, we can substitute the known values:
v = 15.56 m/s + (-5.02 m/s^2) * 2 s
Simplifying the equation:
v = 15.56 m/s - 10.04 m/s
v ≈ 5.52 m/s
Therefore, the car's speed at the moment of impact is approximately 5.52 m/s.
However, it's important to note that the assumption made in this problem is that the car comes to a stop at the barrier. This assumption may not always hold true in real-world situations, so it's always good to double-check and verify the given conditions.