Question

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?

Answer 1

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

Acceleration of the sprinter

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

Distance traveled by the sprinter

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

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