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A particle moves a distance 'x' in time 't' according to equation x=(t+5)^-1 . The acceleration of the particle is propotional to ~ 1.) (velocity)^3/2 2.) (distance)^2 3.) (distance)^-2 4.) (velocity)^2/3 the answer is option 1 . please your justify the answer!
Most Upvoted Answer
A particle moves a distance 'x' in time 't' according to equation x=(t...
Acceleration is the rate of change of velocity with respect to time. In order to determine the acceleration of the particle, we need to find the velocity function first.
1. Finding the velocity function:
We are given the equation for the distance traveled by the particle, x=(t^5)^-1. To find the velocity function, we need to differentiate x with respect to time (t).
Taking the derivative of x with respect to t, we can use the chain rule:
dx/dt = d/dt((t^5)^-1)
= -1(t^5)^-2 * d/dt(t^5)
= -1(t^5)^-2 * 5t^4
= -5t^4/(t^5)^2
= -5t^4/t^10
= -5/t^6
Therefore, the velocity function v(t) = dx/dt = -5/t^6.
2. Finding the acceleration function:
To find the acceleration function, we need to differentiate the velocity function with respect to time (t).
Taking the derivative of v(t) with respect to t:
dv/dt = d/dt(-5/t^6)
= 5(6/t^7)
= 30/t^7
Therefore, the acceleration function a(t) = dv/dt = 30/t^7.
3. Determining the relationship between acceleration and velocity:
Now, let's examine the relationship between acceleration and velocity. We have the velocity function v(t) = -5/t^6 and the acceleration function a(t) = 30/t^7.
Let's substitute the velocity function into the acceleration function:
a(t) = 30/t^7
= 30/((-5/t^6)^7)
= 30/((-5)^7/t^(6*7))
= 30/((-5)^7/t^42)
= 30/(5^7/t^42)
= (30*t^42)/(5^7)
From the above equation, we can see that the acceleration is proportional to (velocity)^3/2. Therefore, the correct option is 1.) (velocity)^3/2.
1. Finding the velocity function:
We are given the equation for the distance traveled by the particle, x=(t^5)^-1. To find the velocity function, we need to differentiate x with respect to time (t).
Taking the derivative of x with respect to t, we can use the chain rule:
dx/dt = d/dt((t^5)^-1)
= -1(t^5)^-2 * d/dt(t^5)
= -1(t^5)^-2 * 5t^4
= -5t^4/(t^5)^2
= -5t^4/t^10
= -5/t^6
Therefore, the velocity function v(t) = dx/dt = -5/t^6.
2. Finding the acceleration function:
To find the acceleration function, we need to differentiate the velocity function with respect to time (t).
Taking the derivative of v(t) with respect to t:
dv/dt = d/dt(-5/t^6)
= 5(6/t^7)
= 30/t^7
Therefore, the acceleration function a(t) = dv/dt = 30/t^7.
3. Determining the relationship between acceleration and velocity:
Now, let's examine the relationship between acceleration and velocity. We have the velocity function v(t) = -5/t^6 and the acceleration function a(t) = 30/t^7.
Let's substitute the velocity function into the acceleration function:
a(t) = 30/t^7
= 30/((-5/t^6)^7)
= 30/((-5)^7/t^(6*7))
= 30/((-5)^7/t^42)
= 30/(5^7/t^42)
= (30*t^42)/(5^7)
From the above equation, we can see that the acceleration is proportional to (velocity)^3/2. Therefore, the correct option is 1.) (velocity)^3/2.
Community Answer
A particle moves a distance 'x' in time 't' according to equation x=(t...
x=( t+5)^-1 . v= dx/dt = -- ( t+5)^ -2. tht means (t+5)= --v ^ -1/2 . now, a= dv/dt = 2 ( t+5) ^ -3. , a= 2( - v ^ -1/2) ^ -3, a = -- 2 v ^ 3/2, tht is a is prop to vel ^3/2
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A particle moves a distance 'x' in time 't' according to equation x=(t+5)^-1 . The acceleration of the particle is propotional to ~ 1.) (velocity)^3/2 2.) (distance)^2 3.) (distance)^-2 4.) (velocity)^2/3 the answer is option 1 . please your justify the answer!
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A particle moves a distance 'x' in time 't' according to equation x=(t+5)^-1 . The acceleration of the particle is propotional to ~ 1.) (velocity)^3/2 2.) (distance)^2 3.) (distance)^-2 4.) (velocity)^2/3 the answer is option 1 . please your justify the answer! for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle moves a distance 'x' in time 't' according to equation x=(t+5)^-1 . The acceleration of the particle is propotional to ~ 1.) (velocity)^3/2 2.) (distance)^2 3.) (distance)^-2 4.) (velocity)^2/3 the answer is option 1 . please your justify the answer! covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves a distance 'x' in time 't' according to equation x=(t+5)^-1 . The acceleration of the particle is propotional to ~ 1.) (velocity)^3/2 2.) (distance)^2 3.) (distance)^-2 4.) (velocity)^2/3 the answer is option 1 . please your justify the answer!.
A particle moves a distance 'x' in time 't' according to equation x=(t+5)^-1 . The acceleration of the particle is propotional to ~ 1.) (velocity)^3/2 2.) (distance)^2 3.) (distance)^-2 4.) (velocity)^2/3 the answer is option 1 . please your justify the answer! for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle moves a distance 'x' in time 't' according to equation x=(t+5)^-1 . The acceleration of the particle is propotional to ~ 1.) (velocity)^3/2 2.) (distance)^2 3.) (distance)^-2 4.) (velocity)^2/3 the answer is option 1 . please your justify the answer! covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves a distance 'x' in time 't' according to equation x=(t+5)^-1 . The acceleration of the particle is propotional to ~ 1.) (velocity)^3/2 2.) (distance)^2 3.) (distance)^-2 4.) (velocity)^2/3 the answer is option 1 . please your justify the answer!.
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