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A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?

Respuesta :

cjrodriguez89

Answer:

Δx = 39.1 m

Explanation:

  • Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:

        [tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]

        where  vf is the final velocity (0 in our case), v₀ is the initial velocity

        (25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

        traveled since the brakes are applied.

  • Solving (1) for Δx, we have:

        [tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]        

asuwafohjames

The car will travel a distance of 39.1 m before its stops.

To solve the problem above, use the equations of motion below.

Equation:

  • v² = u²+2as................... Equation 1

Where:

  • v = final velocity of the automobile
  • u = initial velocity of the automobile
  • a = acceleration
  • s = distance covered

From the question,

Given:

  • v = 0 m/s (before its stops)
  • u = 25 m/s
  • a = -8 m/s² (decelerating)

Substitute these values into equation 1

  • ⇒ 0² = 25²+2(-8)(s)

Solve for s

  • ⇒ 0²-25² = -16s
  • ⇒ -16s = -625
  • ⇒ s = -625/16
  • ⇒ s = 39.1 m

Hence, The car will travel a distance of 39.1 m before its stops.

Learn more about acceleration here: https://brainly.com/question/605631

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