Starting simultaneously from the same spot, O, Alvin and Bernard set out across a field in the gog. Alvin travels with velocity i + 2 j m/s and Bernard with velocity 3i + j m/s, To avoid losing each other, they hold opposite ends of 90 m string.
a) At what speed alvin travels
b) Find the direction of travel of the two people and hence the angle between their path
c) Find the position vector OA and OB at t sesonds, and hence the vector AB at that time
d)Find the value of t for which the dtring become taut
Alvin speed = sqrt (1^2 + 2^2) = sqrt (5)
tan Alvin = 2/1 so angle Alvin = 63.4 deg
tan Bernard = 1/3 so angle = 18.4
so
angle between = 63.4 - 18.4 = 45 degrees
OA = t i + 2t j
OB = 3t i + t j
AB = 2 t i -1 t j
d = sqrt ( 4t^2 + t^2) = t sqrt (5)
90 = t sqrt 5
t = 40.3 seconds
a) To find the speed at which Alvin travels, we need to calculate the magnitude of his velocity vector. The magnitude of a vector (V) can be found using the formula:
|V| = √(Vx^2 + Vy^2)
Given that Alvin's velocity vector is i + 2j m/s, we can substitute the values into the formula:
|Alvin's velocity| = √((1)^2 + (2)^2)
= √(1 + 4)
= √5
Therefore, Alvin travels at a speed of √5 m/s.
b) To find the direction of travel of Alvin and Bernard, we need to calculate the angle between their paths. The angle between two vectors can be found using the dot product:
θ = cos^(-1)((Alvin's velocity) dot (Bernard's velocity)) / (|Alvin's velocity| * |Bernard's velocity|)
Given that Alvin's velocity vector is i + 2j m/s and Bernard's velocity vector is 3i + j m/s, we can substitute the values into the formula:
θ = cos^(-1)(((1)(3) + (2)(1)) / (√5 √10))
= cos^(-1)((3 + 2) / (√5 √10))
= cos^(-1)(5 / (√50))
≈ 49.54 degrees
Therefore, the angle between their paths is approximately 49.54 degrees.
c) To find the position vector OA and OB at time t, we can use the formula:
Position vector = initial position vector + velocity vector * t
Given that the initial position vector O is at the origin (0,0) and Alvin's velocity vector is i + 2j m/s, we can calculate OA:
OA = (0)i + (0)j + (1)i + (2)j * t
= i + 2j * t
Similarly, given that Bernard's velocity vector is 3i + j m/s, we can calculate OB:
OB = (0)i + (0)j + (3)i + (1)j * t
= 3i + j * t
To find the vector AB at time t, we subtract OB from OA:
AB = OA - OB
= (i + 2j * t) - (3i + j * t)
= i + 2j * t - 3i - j * t
= -2i + j * t
Therefore, the vector AB at time t is -2i + j * t.
d) To find the value of t for which the string becomes taut, we need to find the time at which the distance between Alvin and Bernard is equal to the length of the string (90m).
The distance between two points with position vectors A and B can be found using the formula:
Distance = |A - B|
Given that A is the position vector OA at time t and B is the position vector OB at time t, we can substitute their values into the formula:
|AB| = |-2i + j * t|
= √((-2)^2 + (t)^2)
= √(4 + t^2)
We want this distance to be equal to 90m:
√(4 + t^2) = 90
To solve this equation for t, we need to square both sides:
4 + t^2 = 90^2
t^2 = 90^2 - 4
t^2 = 8100 - 4
t^2 = 8096
t = √8096
t ≈ 89.93
Therefore, the string becomes taut at approximately t = 89.93 seconds.