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'm not entirely sure either. I can see that the average velocity v_avg=s/t or 24m/2sec
Is there some formula for average velocity in your text?
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...
You are on the right track! The given formulas are indeed correct and can be used to solve the problem. Let's go step by step:
a) To find the car's constant deceleration before the impact, we can use the equation x-x0 = v0t + 1/2at^2, where x is the final position of the car, x0 is the initial position (distance from the barrier), v0 is the initial velocity (56 km/h converted to m/s), t is the time taken to reach the barrier (2 sec), and a is the constant deceleration we need to find.
Rearranging the equation, we get 1/2at^2 + v0t + x0 - x = 0. Plugging in the values, we have 1/2a(2^2) + 56(2) + 24 - 0 = 0. Simplifying, we obtain 2a + 112 + 24 = 0. Combining like terms, we get 2a + 136 = 0. Subtracting 136 from both sides, we have 2a = -136. Dividing both sides by 2, the constant deceleration is a = -68 m/s^2.
b) To find the car's speed at impact, we can use the equation v = v0 + at, where v is the final velocity we need to find, v0 is the initial velocity (56 km/h converted to m/s), t is the time taken to reach the barrier (2 sec), and a is the constant deceleration (-68 m/s^2).
Plugging in the values, we have v = 56 + (-68)(2). Simplifying, we obtain v = 56 - 136. Combining like terms, the car's speed at impact is v = -80 m/s.
It is important to note that the negative sign indicates that the car is moving in the opposite direction (decelerating) compared to its initial velocity. Since velocity is a vector quantity, negative values indicate opposite directions.