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In a historical movie, two knights on horseback start from rest 86 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 0.26 m/s2. Relative to Sir George's starting point, where do the knights collide?

User Xxyyxx
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Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Step-by-step explanation:

Let the distance covered by Sir George be
S_(1)

and the distance covered by Sir Alfred be
S_(2)

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion


S = ut + (1)/(2)at^(2)

Where
S is the distance traveled


u is the initial velocity


a is the acceleration

and
t is the time

For Sir George,


S = S_(1)


u = 0 m/s (Since they start from rest)


a =0.21 m/s²

Hence,


S = ut + (1)/(2)at^(2) becomes


S_(1) = (0)t + (1)/(2)(0.21)t^(2)\\S_(1) = 0.105 t^(2)\\


t^(2) = (S_(1))/(0.105)

Now, for Sir Alfred


S = S_(2)


u = 0 m/s (Since they start from rest)


a =0.26 m/s²

Hence,


S = ut + (1)/(2)at^(2) becomes


S_(2) = (0)t + (1)/(2)(0.26)t^(2)\\S_(2) = 0.13 t^(2)\\


t^(2) = (S_(2))/(0.13)

Since, they traveled for the same time,
t just before collision, we can write


(S_(1))/(0.105)= (S_(2))/(0.13)

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that


S_(1) + S_(2) = 86 m

Then,
S_(2) = 86 - S_(1)

Then,


(S_(1))/(0.105)= (S_(2))/(0.13) becomes


(S_(1))/(0.105)= (86 -S_(1))/(0.13)


0.13{S_(1)}= 0.105({86 -S_(1)})\\0.13{S_(1)}= 9.03 - 0.105S_(1)}\\0.13{S_(1)} + 0.105S_(1)}= 9.03 \\0.235{S_(1)} = 9.03\\{S_(1)} =(9.03)/(0.235)


S_(1) = 38.43 m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

User Drewjoh
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