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Kazeer [188]
1 year ago
14

When starting a foot race, a 64 kilogram sprinter exerts an average force of 693 newtons backward on the ground for 0.59 seconds

. how far does he travel in meters during this time?
Physics
1 answer:
cluponka [151]1 year ago
6 0

The distance traveled by the sprinter in meters is determined as 1.88 m.

<h3>Acceleration of the sprinter</h3>

The acceleration of the sprinter is the rate of change of velocity of the sprinter with time.

The acceleration of the sprinter is calculated as follows;

Apply Newton's second law of motion as follows;

F = ma

a = F/m

where;

  • F is the applied force by the sprinter
  • m is mass of the sprinter
  • a is acceleration of the sprinter

a = 693 N / 64 kg

a = 10.83 m/s²

<h3>Distance traveled by the sprinter</h3>

The distance traveled by the sprinter is calculated as follows;

s = ut + ¹/₂at²

where;

  • u is initial velocity = 0

s = ¹/₂at²

where;

  • t is time of motion
  • a is acceleration

s = (0.5)(10.83)(0.59²)

s = 1.88 m

Thus, the distance traveled by the sprinter in meters is determined as 1.88 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

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