Question

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 nearest 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? _____________________

Answer 1

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

$d = vt$

$d = 10 (2.5)$

$d = 25 yards$

Answer 2

The time the two players will collide is 2.5 s

The distance traveled by the receiver before the collision is 25 yards.

The given parameters;

• distance the two players, = 40 yards

• velocity of the receiver, Vc = 10 yards/s

• velocity of the tackler, Vt = 6 yards/s

To find:

• the time before they collide

• the distance traveled by the receiver

The time taken before the two players collide is calculated by applying relative velocity principle;

• Let the time when they collide = t

• assume left to right = positive

• right to left = negative

$(V_c - (-V_t))\times t = 40 \ yards$

$(10 + 6)t = 40\\\\16t = 40\\\\t = \frac{40}{16} \\\\t = 2.5 \ seconds$

The time the two players will collide is 2.5 s

The distance traveled by the receiver before the collision is calculated as;

$distance = speed \times time\\\\distance = V_c \times t\\\\distance = 10 \times 2.5= 25 \ yards$

Thus, the distance traveled by the receiver before the collision is 25 yards.

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