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3 years ago

7

In a football game, one team kicks off to the other. At the moment the receiver catches the ball, he is 40 yards from the neares

t tackler. The receiver runs left to right at a speed of 10 yards per second (10 y/s). The tackler runs right to left at a speed of 6 yards per second.
How long does it take before they collide? _____

How far does the receiver go? _____________________

2 answers:

Talja [164]3 years ago

8 0

The answer is hopefully if i'm correct is 5 or 6 seconds

Nikitich [7]3 years ago

6 0

Answer:

Part a)

$\delta t = 2.5 s$

Part b)

$d = 25 yards$

Explanation:

Speed of the receiver is 10 yards per second towards right

Speed of the tackler is 6 yards per second towards left

So here we have net relative speed of the two towards each other is given as

$v_{12} = v_1 - v_2$

$v_{12} = 10 - (-6) = 16 yards/s$

now we know that the distance between receiver and tackler initially is 40 Yards

Part a)

so the time taken by two to collide is given as

$\delta t = \frac{\Delta x}{v}$

$\delta t = \frac{40}{16}$

$\delta t = 2.5 s$

Part b)

Distance moved by the receiver in t = 2.5 s

$d = vt$

$d = 10 (2.5)$

$d = 25 yards$

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where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

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k can cancel out from both side of equation (3), so that we finally obtain the following;

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$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

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3 years ago

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Answer:

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this can be gotten by summing all the distances the car has travelled.

i,e total distance= 60 miles+120 miles

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the displacement can be gotten by subtracting the final distance from the initial distance

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ivolga24 [154]

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