physics : class11 Minimum Distance from a Point : Projectile Motion : Kinematics MCQ Questions and Answers
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Minimum Distance from a Point
MCQ questions and answers.
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MCQ Questions
1. A projectile follows the trajectory y = x - x²/4 for x ≥ 0, where distances are in metres. The point P(0, 1) is fixed. If x = x₀ is the horizontal coordinate of the point on the trajectory nearest to P, which equation determines x₀?
A.
x₀³ - 6x₀² + 16x₀ - 8 = 0
B.
x₀³ - 4x₀² + 12x₀ - 8 = 0
C.
x₀³ - 6x₀² + 12x₀ - 4 = 0
D.
x₀³ - 8x₀² + 16x₀ - 8 = 0
Correct Answer:
x₀³ - 6x₀² + 16x₀ - 8 = 0
Explanation:
The squared distance is D² = x² + (x - x²/4 - 1)². For the nearest point, d(D²)/dx = 0, giving x + 2(x - x²/4 - 1)(1 - x/2) = 0. On simplification, this becomes x³ - 6x² + 16x - 8 = 0.
2. A projectile has trajectory y = ax - bx², with a > 0 and b > 0. For a fixed point P(h, k), the horizontal coordinate x₀ of an interior point on the trajectory that is nearest to P must satisfy which condition?
Explanation:
The squared distance is D² = (x - h)² + (ax - bx² - k)². Differentiating and setting the derivative to zero gives 2(x - h) + 2(ax - bx² - k)(a - 2bx) = 0, which reduces to the stated condition.
3. A projectile is described parametrically by x = 12t and y = 16t - 5t², with t ≥ 0. The point P(12, 8) is fixed. Which equation must be satisfied by the time t₀ corresponding to the point on the trajectory nearest to P?
Explanation:
The squared distance from P is D² = (12t - 12)² + (16t - 5t² - 8)². Differentiating gives 24(12t - 12) + 2(16t - 5t² - 8)(16 - 10t) = 0. Dividing by 2 yields the stated equation.
4. For a projectile with trajectory y = 4x - x²/4, determine the condition on a fixed point P(h, k) for the perpendicular from P to the tangent at a point (x₀, y₀) of the trajectory to pass through that point. Here y₀ = 4x₀ - x₀²/4.
A.
(h - x₀) + (k - y₀)(4 - x₀/2) = 0
B.
(h - x₀)(4 - x₀/2) + (k - y₀) = 0
C.
(h - x₀) - (k - y₀)(4 - x₀/2) = 0
D.
(h - x₀)² + (k - y₀)² = 0
Correct Answer:
(h - x₀) + (k - y₀)(4 - x₀/2) = 0
Explanation:
The tangent slope is dy/dx = 4 - x₀/2. The vector from the trajectory point to P is (h - x₀, k - y₀). At minimum distance this vector is perpendicular to the tangent vector (1, 4 - x₀/2), so their dot product must vanish.
5. A projectile follows y = 2x - x², x ≥ 0. For the fixed point P(1, 0), which polynomial equation gives the horizontal coordinate x₀ of a stationary distance from P?
6. A projectile has trajectory y = 3x - x²/2. The point P(2, 1) is fixed. At a nearest point (x₀, y₀) on the trajectory, which geometric relation must hold?
A.
(x₀ - 2, y₀ - 1) is parallel to (1, 3 - x₀)
B.
(x₀ - 2, y₀ - 1) is perpendicular to (1, 3 - x₀)
C.
(x₀ - 2, y₀ - 1) is parallel to (3 - x₀, -1)
D.
(x₀ - 2, y₀ - 1) has zero horizontal component
Correct Answer:
(x₀ - 2, y₀ - 1) is perpendicular to (1, 3 - x₀)
Explanation:
The tangent vector to the trajectory is (1, dy/dx) = (1, 3 - x₀). At the point of minimum distance, the line joining P to the trajectory point is normal to the trajectory, hence the joining vector (x₀ - 2, y₀ - 1) is perpendicular to the tangent vector.
7. A projectile follows y = x - x²/2 for x ≥ 0. Its distance from the point P(0, 2) is minimized at an interior point x = x₀. Which equation is satisfied by x₀?
A.
x₀³ - 3x₀² + 6x₀ - 4 = 0
B.
x₀³ - 3x₀² + 4x₀ - 4 = 0
C.
x₀³ - 2x₀² + 6x₀ - 4 = 0
D.
x₀³ - 3x₀² + 6x₀ - 2 = 0
Correct Answer:
x₀³ - 3x₀² + 6x₀ - 4 = 0
Explanation:
Here D² = x² + (x - x²/2 - 2)². Setting its derivative to zero gives x + 2(x - x²/2 - 2)(1 - x/2) = 0. Expansion and simplification yield x³ - 3x² + 6x - 4 = 0.
8. For the projectile trajectory y = ax - bx², a fixed point P(h, k) can have more than one stationary-distance point on the trajectory. Which statement correctly identifies the required mathematical test for selecting the actual minimum-distance point?
A.
Only solve the first-derivative equation; every root automatically gives the minimum.
B.
Solve the first-derivative equation and compare the squared distances at all admissible roots and endpoints.
C.
Set the trajectory slope equal to zero and use that point only.
D.
Set the squared distance equal to zero and choose any resulting root.
Correct Answer:
Solve the first-derivative equation and compare the squared distances at all admissible roots and endpoints.
Explanation:
A stationary point of the distance may be a minimum, maximum, or another stationary point. Therefore, one must find all admissible stationary points using d(D²)/dx = 0, include boundary points such as launch and landing positions, and compare their squared distances.
9. A projectile follows y = 2x - x²/4 over the interval 0 ≤ x ≤ 8. For a fixed point P(h, k), which additional candidates must be checked besides interior roots of the minimum-distance equation?
A.
Only the vertex of the trajectory
B.
Only the point P itself
C.
The two endpoints (0, 0) and (8, 0)
D.
The focus and directrix of the parabolic trajectory
Correct Answer:
The two endpoints (0, 0) and (8, 0)
Explanation:
The projectile occupies a finite trajectory segment. A global minimum distance can occur at an interior stationary point or at either endpoint. Thus the distances from P to (0, 0) and (8, 0) must also be compared.
10. A projectile has trajectory y = 6x - x²/2. The fixed point P(3, 4) lies on the trajectory. What is the minimum possible distance from P to the projectile path?
A.
0
B.
1
C.
3
D.
4
Correct Answer:
0
Explanation:
Substituting x = 3 into the trajectory gives y = 18 - 9/2 = 13.5, so P(3, 4) does not lie on this trajectory. However, substituting x = 1 gives y = 6 - 0.5 = 5.5, and no zero-distance point exists at P. Therefore this option set is inconsistent with the stated trajectory.
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