Constant acceleration and collisions?

Question

In an historical movie, two knights on horseback start from rest 74.3 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.264 m/s2, while Sir Alfred's has a magnitude of 0.300 m/s2. Relative to Sir George's starting point, where do the knights collide?

Answer 1

I assume that they both start riding at the same time. If one starts before the other, this procedure can be modified slightly to find the position the earlier one arrives at at the time the later one starts.

Setup a coordinate axis with Sir George at x=0 and Sir Alfred at x=74.3.

If you have a constant acceleration, you can easily derive a formula for the position of either knight after any given time t. We know that acceleration is the derivative of velocity. Let's call velocity v. Then,

dv/dt = a

where a is the acceleration. Move the dt over and take the integral of the right side from 0 to t and integrate the left from an initial velocity to the velocity at time t. The result is an expression for the velocity at time t (and note that velocity is just the derivative of position, which we'll call x):

dx/dt = v(t) = at + v0

where v0 is the initial velocity. Again, bring the dt to the right side and integrate from 0 to t and integrate the left side from an initial position to the position at time t. The result is what we want, an expression for position at time t:

x(t) = 0.5*a*t^2 + v0*t + x0

where x0 is the initial position.

So Sir George's position is expressed by:

xG(t) = 0.5*(0.264)*t^2 + 0*t + 0 = 0.132*t^2

and Sir Alfred's position is expressed by:

xA(t) = 0.5*( -0.3 )*t^2 + 0*t + 74.3 = -0.15*t^2 + 74.3

You would like to find the time (which will give you position) where the two of them are equal. Thus, set the two position expressions equal to each other and solve for time:

0.132*t^2 = -0.15*t^2 + 74.3

0.282*t^2 = 74.3

t^2 = 74.3/0.282 = 263.475

t = 16.232

Now plug that into either expression (you can also use the expression for v(t) above to find the velocity of each of them at 16.232 seconds):

xG(t) = 34.779

xA(t) = 34.779


So your answers are:

*) Thus, they collide at 34.779 meters away from Sir George's starting point.

*) This happens 16.232 seconds after the two of them start.

*) When they collide, Sir George is traveling 4.285 m/s toward Sir Alfred, and Sir Alfred is traveling 4.870 m/s toward Sir George.

Answer 2

I have a much simpler solution
let's assume that the distance of collision from Sir George's position is "x" m
then the distance from Sir Alfred's position will be "74.3-x"m
(Since the total distance in between them is 74.3m)
Now for let the collision occur after "t" secs
then for Sir G
u(1)=0 m/s
a(1)=0.264m/s^2
t(1)=tsec
then S(1)=u(1)*t(1)+ 1/2* a(1)*t^2
x= 0+ 1/2*0.264*t^2
x=0.132t^2 -------eqn 1

For Sir A
u(2)=0m/s
a(2)=0.300m/s^2
t(2)=t sec
then
S(2)= u(2)*t(2)+1/2*a(2)*t(2)^2
74.3-x=0+1/2*0.300*t^2
74.3-x=0.15t^2 --------------- eqn 2

adding eqn 1 and 2
74.3-x=0.150t^2
x=0.132t^2
-----------------------------
74.3 =0.282t^2
74.3/0.282=t^2
263.4751=t^2
on taking square root
t=16.2319sec

putting this value in eqn 1
x=0.132t^2
x=0.132*263.4751
x=34.779 m
so the collision will take place at a distance of 34.779m from Sir G's positiom after an interval of 16.2319 sec

Answer 3

The knights will take the same time to collide.
The total Distance traveled is =74.3 m
The toal acceleration of aproach =..264+.300=.564
The total time traveled by each knight =[74.3/.564]^.5=11.47 sec
Collision distance of sir george =.264x(11.47)^2=34.778meters.
if the mass of the knight are the same the total collision acceleration is =.3996198193 meter/sec^2

Answer 4

To the right.