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saul85 [17]
3 years ago
9

A soccer player practices kicking the ball into the goal from halfway down the soccer field. The time it takes for the ball to g

et to the goal from the time it is kicked averages about three seconds. The soccer field is 90 meters long. The player wants to know the average velocity of the ball. Is there enough information to calculate this?
A) No, only average speed can be calculated from the distance of 90 meters and the time of three seconds.

B) Yes, displacement is 45 meters, elapsed time is three seconds, and the direction is toward the goal.

C) No, only average speed can be calculated from the distance of 45 meters and the time of three seconds.

D) Yes, displacement is 90 meters, elapsed time is three seconds, and the direction is toward the goal.
Physics
2 answers:
12345 [234]3 years ago
4 0

This is correct, I just did the test. Yes, displacement is 45 meters, elapsed time is three seconds, and the direction is toward the goal.

Effectus [21]3 years ago
3 0

Answer:

b

Explanation:

i took the test

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If 10 waves pass a point each second and their wavelength is 30m, what is their speed?
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On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
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Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
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