Question #77336
2 Answers
Answer:
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
#Deltax = ((v_(0x) + v_x)/2)t#
to find the distance traveled,
Here,

#v_x# is the final velocity (i.e. their maximum speed) 
#v_(0x)# 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.
We're given that the first knight has a max speed of
#Deltax = ((0 + 18.0color(white)(l)"m/s")/2)(6.00color(white)(l)"s")#
#Deltax = ul(54color(white)(l)"m"#
We're given that the second knight has a max speed of
#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# #"m"# # 54# #"m"# # 37.5# #"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
So, traveling for
#Deltax = (15.0color(white)(l)"m/s")(1color(white)(l)"s") = ul(15color(white)(l)"m"#
Therefore, the distance between the two knights after
#108.5# #"m"# # 15# #"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

#"knight 1": x = 0# 
#"knight 2": x = 93.5# #"m"#
(distance separating them now is
Knight one will be traveling in the positive
Their position functions are thus
#"knight 1 ": ul(x = (18.0color(white)(l)"m/s")t#
#"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
#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
#(33.0color(white)(l)"m/s")t = 93.5color(white)(l)"m"#
#color(red)(ul(t = 2.83color(white)(l)"s"#
REMEMBER: this is the time it takes after the
#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
Suppose the knights meet
Applicable kinematic expressions
#v=u+at# .....(1)
#s=vt# .....(2)
Knight 1 after dropping of flag.
Acceleration
#18.00=0+a_1xx6.00#
#=>a_1=18.0/6.00=3.00ms^2#
 Distance traveled during acceleration.
#v_1^2u_1^2=2a_1s_1#
#(18.0)^20^2=2xx3.00xxs_1#
#=>s_1=(18.0)^2/6.00#
#=>s_1=54m#  Distance traveled with constant velocity
#s_1^'=v_1xxt_1#
#s_1^'=18.0xx(t6.00)#
Total distance covered
#s_1+s_1^'=54+18t108#
#s_1+s_1^'=18t54# .....(3)
Knight 2 after dropping of flag.
Acceleration
#15.00=0+a_2xx5.00#
#=>a_2=15.0/5.00=3.00ms^2#
 Distance traveled during acceleration.
#v_2^2u_2^2=2a_2s_2#
#(15.0)^20^2=2xx3.00xxs_2#
#=>s_2=(15.0)^2/6.00#
#=>s_2=37.5m#  Distance traveled with constant velocity
#s_2^'=v_2xxt_2#
#s_2^'=15.0xx(t5.00)#
Total distance covered
#s_2+s_2^'=37.5+15t75#
#s_2+s_2^'=15t37.5# .....(4)
Adding (3) and (4)
#s_1+s_1^'+s_2+s_2^'=18t54+15t37.5=# Distance covered by both knights#=200m#
#18t54+15t37.5=200#
#=>33t=200+54+37.5#
#=>33t=291.5#
#=>t=8.8bar3s# , rounded to one decimal place.