I assume L=120 yards as the length of the football field.
1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
And the time taken is t=22.4 s, so the average speed of the player is
2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
and so, the average velocity is
Answer:
0.11 m/s
Explanation:
From the question given above, the following data were obtained:
Initial displacement (d1) = 1.09 m
Final displacement (d2) = 2.55 m
Time (t) = 12.8 s
Average velocity =?
Next, we shall determine the total displacement (i.e change in displacement). This can be obtained as follow:
Initial displacement (d1) = 1.09 m
Final displacement (d2) = 2.55 m
Total displacement = d2 – d1
Total displacement = 2.55 – 1.09
Total displacement = 1.46 m
Finally, we shall determine the average velocity of the beetle. This can be obtained as follow:
Total Displacement = 1.46 m
Total time (t) = 12.8 s
Average velocity =?
Average velocity = Total Displacement / Total time
Average velocity = 1.46/12.8
Average velocity = 0.11 m/s
Thus, the average velocity of the beetle is 0.11 m/s
Answer:
You can drink 18 drinks and 1.38 dollars left with you.
Explanation:
1 drink cost 1.59 dollars
You have 30 dollars with you.
Number of drinks in 30 dollars = 
Number of drinks in 30 dollars = 18
From division 1.36 remains as remainder so, 1.38 dollars left with you.
Answer:
The cart moved away from the starting point between 8 s and 10 s.
Explanation:
Given that :
Which of these does NOT describe the cart’s motion on this graph?
The cart was at rest between 5 s and 7 s : From the distance of golf cart Vs time graph ; the car was at rest between 5s and 7s as the graph was flat with that time interval, meaning there was no change in distance.
The cart moved toward the starting point between 7 s and 12 s. : The graph depicts a negative slope at this time interval as the distance from starting point fell from about 26 m to 10 m
The cart moved away from the starting point between 8 s and 10 s. : At this time interval, the cart moved towards the starting not and not away. This could be seen in the decrease in Distance from starting point between the tune interval.
The cart moved away from the starting point between 2 s and 5 s. - - > The cart moved away from the starting point, with the positive slope signifying an increase in distance.
Therefore, The cart moved away from the starting point between 8 s and 10 s does not describe the motion of the cart on the graph.
Answer: what do u need help with
Explanation:
The giant-impact hypothesis, sometimes called the Big Splash, or the Theia Impact, suggests that the Moon formed from the ejecta of a collision between the ...
