Acar traveling 56.0 km/h is 24.0 m from a barrier when the driver slams on the brakes. the car hits the barrier 2.00 s later. (a) what is the magnitude of the car’s constant acceleration before impact? (b) how fast is the car traveling at impact?

(A) acceleration = -3.6m/s². This is deceleration or negative acceleration.

(B) Therefore, velocity just before impact, v = 8.4m/s

Explanation:

(a) Initial velocity, u = 56km/h = (56*1000/3600)m/s = 15.6m/s

Distance, s = 24.0m, time, t = 2.0s, acceleration, a = ?

Using the equation, s = ut + 1/2at²

24.0m = 15.6m/s * 2s + (a * 2²)/2

24.0 = 31.2 +2a

2a = -31.2 + 24

2a = -7.2

a = -7.2/2

acceleration = -3.6m/s². This is deceleration or negative acceleration.

(b) final velocity just before impact, v = ? initial velocity, u = 15.6m/s, acceleration = -3.6m/s², time, t = 2.0s

using the equation v = u + at

v = 15.6m/s + (-3.6m/s² * 2s)

v = 15.6m/s - 7.2m/s

v = 8.4m/s

Therefore, velocity just before impact, v = 8.4m/s

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.