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In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?

In The Figure Particle A Moves Along The Line Y 31 M With A Constant Velocity V With Arrow Of Magnitude 28 Ms And Parallel To The X Axis At The Instant Particle class=

Sagot :

oladayoademosu

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

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