Two football players start from rest, 12 m apart. They run directly towards each other, both players accelerating. The first player has an acceleration of 5 m/s^2 and the second has an acceleration of 4 m/s^2. a) how much time passes before they collide? b) at the instant they collide, how far has the first player run?
To find the answers to these questions, we can use the equations of motion. Let's break down the problem step by step.
a) How much time passes before they collide?
We can use the equation of motion:
\[s = ut + \frac{1}{2}at^2\]
where "s" is the distance traveled, "u" is the initial velocity, "a" is the acceleration, and "t" is the time.
Since both players start from rest, their initial velocity ("u") is 0. The first player has an acceleration of 5 m/s^2, and the second player has an acceleration of 4 m/s^2.
Let's calculate the time it takes for the players to collide:
For the first player:
\[s_1 = 0t + \frac{1}{2} \cdot 5 \cdot t^2\]
\[s_1 = \frac{5}{2}t^2\]
For the second player:
\[s_2 = 0t + \frac{1}{2} \cdot 4 \cdot t^2\]
\[s_2 = 2t^2\]
Since they both collide at the same time, we can equate the two distances:
\[12 = \frac{5}{2}t^2 + 2t^2\]
\[12 = \frac{5}{2}t^2 + 2t^2\]
\[12 = \frac{5+4}{2}t^2\]
\[12 = \frac{9}{2}t^2\]
\[\frac{24}{9} = t^2\]
\[t^2 = \frac{24}{9}\]
\[t^2 = \frac{8}{3}\]
To solve for "t", we take the square root of both sides:
\[t = \sqrt{\frac{8}{3}}\]
Hence, the time it takes for the players to collide is approximately \(t \approx 1.632\) seconds.
b) At the instant they collide, how far has the first player run?
To find the distance the first player has run, substitute the time "t" found in part (a) into the equation of motion for the first player:
\[s_1 = u_1t + \frac{1}{2}a_1t^2\]
Since the initial velocity for the first player is 0,
\[s_1 = \frac{1}{2} \cdot 5 \cdot t^2\]
\[s_1 = \frac{5}{2}t^2\]
Plugging the value of "t" we found earlier, we can calculate the distance:
\[s_1 = \frac{5}{2} \cdot 1.632^2\]
Hence, the first player has run approximately \(s_1 \approx 6.829\) meters.