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insens350 [35]
3 years ago
12

The leg's force forward on the foot= 500N

Physics
1 answer:
Anna11 [10]3 years ago
5 0

There's so much going on here, in a short period of time.

<u>Before the kick</u>, as the foot swings toward the ball . . .

-- The net force on the ball is zero.  That's why it just lays there and
does not accelerate in any direction.

-- The net force on the foot is 500N, originating in the leg, causing it to
accelerate toward the ball.


<u>During the kick</u> ... the 0.1 second or so that the foot is in contact with the ball ...

-- The net force on the ball is 500N.  That's what makes it accelerate from
just laying there to taking off on a high arc.

-- The net force on the foot is zero ... 500N from the leg, pointing forward,
and 500N as the reaction force from the ball, pointing backward. 

That's how the leg's speed remains constant ... creating a dent in the ball
until the ball accelerates to match the speed of the foot, and then drawing
out of the dent, as the ball accelerates to exceed the speed of the foot and
draw away from it.


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A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?
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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

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