Two particle are moving along two long straight lines, in the same plane with same
speed equal to 1
20cms
. The angle between the two lines is 0
60 and their intersection
point is O . At a certain moment, the two particles are located at distances 3m and
4m from O and are moving towardsO . Subsequently the shortest distance between
them will be
1) 50cm 2) 40 2cm 3) 50 2cm 4) 50 3cm
Answers
Answer:
time t=0,the first particle is 300cm from O and moving toward O at a velocity of 20cm/s.
The distance of this particle to O as a function of time is:
A=300−20t
At the time t=0, the second particle is 400cm from O and moving toward O at a velocity of 20m/s.
The distance of this particle to O as a function of time is:
B=400−20t
The distance between the 2 particles can be found using the cosine rule
C
2
=A
2
+B
2
−2ABcos60
0
C
2
=(300−20t)
2
+(400−20t)
2
−2(300−20t)(400−20t)cos60
0
C
2
=(400t
2
−12000t+90000)+(400t
2
−16000t+160000)−(400t
2
−14000t+120000)
C
2
=400t
2
−14000t+130000
C=
(400t
2
−14000t+130000)
Now one needs to find t such that C will be minimized.
If C is minimized at a certain t, then also C
2
will be minimized at that t.
So instead of finding the first derivative of C (which is a square root,
so a lot of work), one can simply find the first derivative of C
2
(a
quadratic expression, which is a lot less work)
C
2
=400t
2
−14000t+130000
dt
2
dC
2
=800t−14000
800t−14000=0
t=17.5s
At time t=17.5s the distance between the 2 particles is minimized.
The distance at that time will be:
C=
(400×17.5
2
−14000×17.5+130000)
C=
7500
⇒C=50
3
Answer:
4 ) 5 0 √ 3 c m . . . . . . . . . . . . . .