# Question 77336

Aug 1, 2017

$t = 8.83$ $\text{s}$

#### Explanation:

We're asked to find the time when the two knights meet; i.e. when they have the same position.

Here's how I went about it:

What we can do first is, for each knight, find the distance traveled during the time they are acceleration to their maximum speed.

For both jousters, we can use the kinematics equation

$\Delta x = \left(\frac{{v}_{0 x} + {v}_{x}}{2}\right) t$

to find the distance traveled, $\Delta x$.

Here,

• ${v}_{x}$ is the final velocity (i.e. their maximum speed)

• ${v}_{0 x}$ is the initial velocity (which will be $0$, since they start from a position of rest)

• $t$ is the time it takes to accelerate to the maximum speed.

bb("KNIGHT ONE:"

We're given that the first knight has a max speed of $18.0$ $\text{m/s}$ (final velocity, v_x), and can reach that speed in $6.00$ $\text{s}$ (time, $t$), so we have

Deltax = ((0 + 18.0color(white)(l)"m/s")/2)(6.00color(white)(l)"s")

Deltax = ul(54color(white)(l)"m"

bb("KNIGHT TWO:"

We're given that the second knight has a max speed of $15.0$ $\text{m/s}$ (final velocity, v_x), and can reach that speed in $6.00$ $\text{s}$ (time, $t$), so we have

Deltax = ((0 + 15.0color(white)(l)"m/s")/2)(5.00color(white)(l)"s")

Deltax = ul(37.5color(white)(l)"m"

These values represent the distances each knight traveled toward each other, so the distance between them now is

$200$ $\text{m}$ $- 54$ $\text{m}$ $- 37.5$ $\text{m}$ = ul(108.5color(white)(l)"m"

$- - - - - - - - - - - - - - - - - - - - -$

Now to make this a little easier, we'll make the times be equal to each other (they are currently at $t = 6.00$ $\text{s}$ and $t = 5.00$ $\text{s}$). What we can do is find the new position of the second knight one second after he reaches his top speed (because he reached his top speed one second before knight one).

So, traveling for $1.00$ $\text{s}$ at his max speed of $15.0$ $\text{m/s}$, he would travel a distance

Deltax = (15.0color(white)(l)"m/s")(1color(white)(l)"s") = ul(15color(white)(l)"m"

Therefore, the distance between the two knights after $6.00$ seconds is

$108.5$ $\text{m}$ $- 15$ $\text{m}$ = ul(93.5color(white)(l)"m"

$- - - - - - - - - - - - - - - - - - - - -$

Now that both knights are traveling at a constant speed toward each other, we can call the new positions of the knights

• $\text{knight 1} : x = 0$

• $\text{knight 2} : x = 93.5$ $\text{m}$

(distance separating them now is $93.5$ $\text{m}$, as we've just calculated)

Knight one will be traveling in the positive $x$-direction (toward knight 2), and knight two will be traveling in the negative $x$-direction.

Their position functions are thus

"knight 1 ": ul(x = (18.0color(white)(l)"m/s")t

$\text{knight 2 ": ul(x = (-15.0color(white)(l)"m/s")t + 93.5color(white)(l)"m}$

Finally, to find the time when they meet, we can sfcolor(red)("set these two equations equal to each other", because we're trying to find the time $t$ when their positions are equal:

overbrace((18.0color(white)(l)"m/s")t)^"knight 1" = overbrace((-15.0color(white)(l)"m/s")t + 93.5color(white)(l)"m")^"knight 2"

Now we just solve for $t$:

(33.0color(white)(l)"m/s")t = 93.5color(white)(l)"m"

color(red)(ul(t = 2.83color(white)(l)"s"

bb("BUT WAIT!"

REMEMBER: this is the time it takes after the $6.00$ $\text{s}$ where we started, so the actual time from the beginning is

t = 6.00color(white)(l)"s" + color(red)(2.83color(white)(l)"s") = color(blue)(ulbar(|stackrel(" ")(" "8.83color(white)(l)"s")" "|)

Thus, the two knights will collide after color(blue)(8.83 sfcolor(blue)("seconds"#.

Aug 1, 2017

Suppose the knights meet $t$ second after dropping of the flag.

Applicable kinematic expressions

$v = u + a t$ .....(1)
$s = v t$ .....(2)

Knight 1 after dropping of flag.

Acceleration ${a}_{1}$

$18.00 = 0 + {a}_{1} \times 6.00$
$\implies {a}_{1} = \frac{18.0}{6.00} = 3.00 m {s}^{-} 2$

1. Distance traveled during acceleration.
${v}_{1}^{2} - {u}_{1}^{2} = 2 {a}_{1} {s}_{1}$
${\left(18.0\right)}^{2} - {0}^{2} = 2 \times 3.00 \times {s}_{1}$
$\implies {s}_{1} = {\left(18.0\right)}^{2} / 6.00$
$\implies {s}_{1} = 54 m$
2. Distance traveled with constant velocity
${s}_{1}^{'} = {v}_{1} \times {t}_{1}$
${s}_{1}^{'} = 18.0 \times \left(t - 6.00\right)$
Total distance covered
${s}_{1} + {s}_{1}^{'} = 54 + 18 t - 108$
${s}_{1} + {s}_{1}^{'} = 18 t - 54$ .....(3)

Knight 2 after dropping of flag.

Acceleration ${a}_{2}$

$15.00 = 0 + {a}_{2} \times 5.00$
$\implies {a}_{2} = \frac{15.0}{5.00} = 3.00 m {s}^{-} 2$

1. Distance traveled during acceleration.
${v}_{2}^{2} - {u}_{2}^{2} = 2 {a}_{2} {s}_{2}$
${\left(15.0\right)}^{2} - {0}^{2} = 2 \times 3.00 \times {s}_{2}$
$\implies {s}_{2} = {\left(15.0\right)}^{2} / 6.00$
$\implies {s}_{2} = 37.5 m$
2. Distance traveled with constant velocity
${s}_{2}^{'} = {v}_{2} \times {t}_{2}$
${s}_{2}^{'} = 15.0 \times \left(t - 5.00\right)$
Total distance covered
${s}_{2} + {s}_{2}^{'} = 37.5 + 15 t - 75$
${s}_{2} + {s}_{2}^{'} = 15 t - 37.5$ .....(4)

${s}_{1} + {s}_{1}^{'} + {s}_{2} + {s}_{2}^{'} = 18 t - 54 + 15 t - 37.5 =$ Distance covered by both knights $= 200 m$
$18 t - 54 + 15 t - 37.5 = 200$
$\implies 33 t = 200 + 54 + 37.5$
$\implies 33 t = 291.5$
$\implies t = 8.8 \overline{3} s$, rounded to one decimal place.