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Elanso [62]
3 years ago
7

a soccer player is running at 6m/s. he then stumbles over an opponents foot falling, and rolling to a stop. this took 4 seconds.

what was his acceleration?
Physics
2 answers:
Natali [406]3 years ago
8 0

a =  \frac{(vf - vi)}{t}  \\ a =  \frac{0 - 6}{4} \\ a =  - 1.5 \frac{m}{ {s}^{2} }
top one is the formula
mid- i plug in the numbers
bottom- answer
marishachu [46]3 years ago
5 0

Answer:

The acceleration of the soccer player is 1.5\ m/s^2 and it is decelerating.

Explanation:

Given that,

Initial speed of a soccer player, u = 6 m/s

Finally, it stops, v = 0

Time, t = 4 s

We need to find the acceleration of a soccer player. We know that the rate of change of velocity is called acceleration of an object. It is given by :

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-6}{4}\\\\a=-1.5\ m/s^2

So, the acceleration of the soccer player is 1.5\ m/s^2 and it is decelerating.

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Answer:

<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>

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Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction  opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move  can be calculated using equation 1.

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