A 2 meter long object is fired vertically upwards from the mid point of two locations A and B, 8 meters apart. The speed of the object after t seconds is given by ds/dt = (2t + 1) meter per second. Let α and β be the angles subtended by the object A and B, respectively after one and two seconds. Find the value of cos (α – β).

Jitender Pal
8 years ago
Hello Student,

Let O be the mid point of AB = 8 m
∴ OA = OB = 4 m
Also OP = 2 m is the initial position of the object which is 2m long. Also we are given, ds/dt = 2t + 1 Integrating we get, s = t2 + t + k (where s is the distance of top of object from O)
When t = 0, s = OP = 2
∴ k = 2
∴ s = t2 + t + 2 . . . . . . . . . . . . . . . . . . . . . . . (1)
For t = 1, s = 4 = OQ ∴ PQ = 2 where PQ is the position of object after 1 sec.
For t = 2, s = 8 = OS but RS = 2 where RS is the position of the object after 2 sec.
∴ OR = OS – RS = 6
Also, OQ = 4
∴ QR = OR – OQ = 6 – 4 = 2
∴ OP = PQ = QR = RS = 2
As per the condition of the question, PQ and RS, the position of the object at t = 1 and t = 2 subtend angles α and β at A and B respectively.
Now let ∠ PAO = θ so that tan θ = 2/4 = 1/2
Also tan (α + θ) = 4/4 = 1
Now, tan α = tan [(α + θ) – θ]
= tan (α + θ) – tan θ/1 + tan (α + θ) tan θ = $\frac{1-\frac{1}{2}}{1+\frac{1}{2}}$= 1/3
∴ sin α = 1/√10 and cos α = 3/√10 . . . . . . . . . . . . . . . . . . . . . (2)
Similarly taking ∠ RAQ = ? so that
tan β = tan[ (θ + α + ? + β) – (θ + α + β)]
= tan (θ + α + ? + β) – tan θ + α + β)/1 + tan (θ + α + ? + β). tan (θ + α + β)
= $\frac{\frac{8}{4}-\frac{6}{4}}{1+\frac{8}{4}x\frac{6}{4}}$= 2-3/2 / 1+ 2(3/2) = 1/8
∴ sin β = 1/√65 and cos β = 8/√65 . . . . . . . . . . . . . . . . . . . . . . . (3)
∴ cos (α – β) = cos α cos β + sin α sin β Using equations (2) and (3) we get
Cos (α – β) = 3/√10. 8/√65 + 1/√10. 1/√65
= 25/5 √2 √13 = 5/√26
Thanks
Jitender Pal