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What is the shortest stopping time for a car which is travelling at 16 m s⁻¹ and has a maximum deceleration of 2.5 m s⁻²? - Leaving Cert Physics - Question (a) - 2013

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Question (a)

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What is the shortest stopping time for a car which is travelling at 16 m s⁻¹ and has a maximum deceleration of 2.5 m s⁻²?

Worked Solution & Example Answer:What is the shortest stopping time for a car which is travelling at 16 m s⁻¹ and has a maximum deceleration of 2.5 m s⁻²? - Leaving Cert Physics - Question (a) - 2013

Step 1

Calculate using the formula for uniform acceleration

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Answer

To find the stopping time, we can use the equation of motion:

v=u+atv = u + at

Where:

  • vv is the final velocity (0 m/s, when the car stops)
  • uu is the initial velocity (16 m/s)
  • aa is the acceleration (deceleration in this case, so it's -2.5 m/s²)

Rearranging the equation gives:
0=162.5t0 = 16 - 2.5t
Solving for tt: 2.5t=162.5t = 16 t=162.5=6.4extst = \frac{16}{2.5} = 6.4 ext{ s}

Step 2

Final answer

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Answer

The shortest stopping time for the car is 6.4 seconds.

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