contestada

A car travelling at 91.0 km/h approaches the turnoff for a restaurant 30.0 m ahead. If the driver slams on the brakes with an acceleration of -6.40 m/s2
2
what will her stopping distance be?

Respuesta :

Eduard22sly

The stopping distance of the car is 49.9 m

Data obatined from the question

The following data were obtained from the question:

  • Initial velocity (u) = 91 Km/h = 91 / 3.6 = 25.28 m/s
  • Acceleration (a) = -6.4 m/s²
  • Final velocity (v) = 0 m//s
  • Stopping distance (s) = ?

How to determine the stopping distance

The stopping distance can be obtained as illustrated below:

v² = u² + 2as

0² = 25.28² + (2 × -6.4 × s)

0 = 639.0784 - 12.8s

Collect like terms

-12.8s = 0 - 639.0784

-12.8s = -639.0784

Divide both sides by -12.8

s = -639.0784 / -12.8

s = 49.9 m

Thus, the stopping distance of the car is 49.9 m

Learn more about velocity:

https://brainly.com/question/3411682

#SPJ1

Otras preguntas