Question

In a historical movie, two knights on horseback start from rest 64 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.21 m/s2, while Sir Alfred's has a magnitude of 0.29 m/s2. Relative to Sir George's starting point, where do the knights collide?

Answer 1

DATA

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$x_{oA}=64 \, m$

$a_{G}=0.21 \, m/s^{2}$

$a_{A}=0.29 \, m/s^{2}$

$x=?$

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SOLUTION

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Set origin at Sir George's starting point. The position of Sir George's as a function of time can be expressed as

$\:$

$x_{G}(t)=\frac{a_{G}t^{2}}{2} \: \: \: \: \: \: \: \: \: \: \: (1)$

$\:$

The position of Sir Alfred's as a function of time can be expressed as

$\:$

$x_{A}(t)=x_{oA}-\frac{a_{A}t^{2}}{2} \: \: \: \: \: \: \: \: \: (2)$

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Let t_c be the collision time, then

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$x_{G}(t=t_{c})=x_{A}(t=t_{c})\equiv x$

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From eq. (1),

$\:$

$x_{G}(t=t_{c})=x=\frac{a_{G}t_{c}^{2}}{2}$

$\Rightarrow \: \: \: t_{c}^{2}=\frac{2x}{a_{G}} \: \: \: \: \: \: \: \: \: \: (3)$

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From eq. (2),

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$x_{A}(t=t_{c})=x=x_{oA}-\frac{a_{A}t_{c}^{2}}{2} \: \: \: \: \: \: \: \: \: \: (4)$

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inserting eq. (3) into (4), we obtain

$\:$

$x=x_{oA}-\frac{a_{A}}{2}\left ( \frac{2x}{a_{G}} \right )$

$\Rightarrow \: \: \: x=x_{oA}- \frac{xa_{A}}{a_{G}}$

$\Rightarrow \: \: \: x+\frac{xa_{A}}{a_{G}}=x_{oA}$

$\Rightarrow \: \: \: x\left (\frac{a_{G}+a_{A}}{a_{G}} \right )=x_{oA}$

$\Rightarrow \: \: \: x=\frac{x_{oA}a_{G}}{a_{G}+a_{A}}$

$\Rightarrow \: \: \: x=\frac{(64m)(0.21 \, m/s^{2})}{0.21 \, m/s^{2}+0.29\, m/s^{2}}$

$\:$

or

$\:$

${\color{Blue} x=26.9 \, m}$

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The knights collide 26.9 m away from the Sir George's starting point.

Answer 2

SOLUTION :

Let they collide after t secs. 

So,

Distance travelled by Sir George = 1/2 a t^2 = 1/2 * 0.21 * t^2 = 0.105 t^2

Distance travelled by Sir Alfred = 1/2 a’ t^2 = 1/2 * 0.29* t^2 = 0.145 t^2

As they collide, the sum of their distances travelled = 91 m

=> 0.105 t^2 + 0.145 t^2 = 64

=> 0.250 t^2 = 64

=> t^2 = 64 / 0.25 = 256 

Sir George ’s distance travelled = 0.105 * 256 = 26.88 m 

Hence,  they collide at a  point 26.88  m  from the starting point of Sir George. (ANSWER).