Acar is traveling at 24.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant acceleration. the car comes to a stop in a distance of 120 m. how fast was the car moving when it was 60.0 m past the point where the brakes were applied?

The answer to your question is : vf = 15.18 m/s

Explanation:

Data

vo = 24 m/s

d = 120 m

vf = ? when d = 60.0 m

Formula

              vf² = vo²  + 2ad

For d =100m

            a = (vf² - vo²) / 2d

            a = (0 -24²) / 2(100)

            a = -576/200

             a = 2.88 m/s²

Now, when d = 60

        vf² = (24)² - 2(2.88)(60)

        vf² = 576 - 345.6

        vf² = 230.4

        vf = 15.18 m/s

17 m/s

Explanation:

Data

initial speed, Vi = 24 m/s

distance travelled, x = 120 m

final speed, Vf = 0 m/s

With help of the next Uniformly Accelerated Motion equation, we can compute car acceleration.

Vf² = Vi² + 2ax

a = Vf² - Vi²/(2x) 

a = 0² - 24²/(2*120) 

a = -2.4 m/s²

Now, the final speed is the unknown and the distance is 60 m:

Vf² = Vi² + 2ax

Vf = √(24² + 2*(-2.4)*60)

Vf = 17 m/s

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.