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A particle moves a distance x in time t according to equation x = (t + 5)–1. The acceleration of particle is proportional to: [2010]
- a)(velocity)3/2
- b)(distance)2
- c)(distance)–2
- d)(velocity)2/3
Correct answer is option 'A'. Can you explain this answer?
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A particle moves a distance x in time t according to equation x = (t +...
We can find the velocity and acceleration of the particle by differentiating the equation for displacement with respect to time:
x = (t - 5)^-1
dx/dt = -(t - 5)^-2 * 1
v = dx/dt = -(t - 5)^-2
d^2x/dt^2 = -(-2)(t - 5)^-3 * 1
a = d^2x/dt^2 = 2(t - 5)^-3
Since we want to find what the acceleration is proportional to, we can write:
a = kx^n
where k is a constant and n is the exponent we want to find.
Substituting the expression for acceleration and displacement, we get:
2(t - 5)^-3 = k(t - 5)^-1
Dividing both sides by (t - 5)^-3, we get:
2 = k(t - 5)^2
Taking the square root of both sides, we get:
√2 = k(t - 5)
Solving for t, we get:
t = √2/k + 5
Substituting this value of t back into the expression for acceleration, we get:
a = 2(√2/k + 5 - 5)^-3
Simplifying, we get:
a = 2/(√2/k)^3
a = 8k^3/√2
Since we want to find what the acceleration is proportional to, we can write:
a ∝ x^n
Substituting the expression for acceleration and displacement, we get:
8k^3/√2 ∝ (t - 5)^-1
8k^3/√2 ∝ x^-1
Taking the reciprocal of both sides, we get:
√2/(8k^3) ∝ x
Therefore, the acceleration of the particle is proportional to the square root of 2 divided by the cube of the constant k, which is proportional to the distance x. So the answer is (c) distance.
x = (t - 5)^-1
dx/dt = -(t - 5)^-2 * 1
v = dx/dt = -(t - 5)^-2
d^2x/dt^2 = -(-2)(t - 5)^-3 * 1
a = d^2x/dt^2 = 2(t - 5)^-3
Since we want to find what the acceleration is proportional to, we can write:
a = kx^n
where k is a constant and n is the exponent we want to find.
Substituting the expression for acceleration and displacement, we get:
2(t - 5)^-3 = k(t - 5)^-1
Dividing both sides by (t - 5)^-3, we get:
2 = k(t - 5)^2
Taking the square root of both sides, we get:
√2 = k(t - 5)
Solving for t, we get:
t = √2/k + 5
Substituting this value of t back into the expression for acceleration, we get:
a = 2(√2/k + 5 - 5)^-3
Simplifying, we get:
a = 2/(√2/k)^3
a = 8k^3/√2
Since we want to find what the acceleration is proportional to, we can write:
a ∝ x^n
Substituting the expression for acceleration and displacement, we get:
8k^3/√2 ∝ (t - 5)^-1
8k^3/√2 ∝ x^-1
Taking the reciprocal of both sides, we get:
√2/(8k^3) ∝ x
Therefore, the acceleration of the particle is proportional to the square root of 2 divided by the cube of the constant k, which is proportional to the distance x. So the answer is (c) distance.
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A particle moves a distance x in time t according to equation x = (t + 5)–1. The acceleration of particle is proportional to: [2010]a)(velocity)3/2b)(distance)2c)(distance)–2d)(velocity)2/3Correct answer is option 'A'. Can you explain this answer?
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