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A particle moves in such a way that its distance from a fixed point,at any instant "t," is given by: x=3t^3 2t^2 5t-9 where "x'" is in centimeters and "t'" in second.What will be its acceleration at the end of "5" second:?
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A particle moves in such a way that its distance from a fixed point,at...
Explanation:
To find the acceleration of the particle after 5 seconds, we need to differentiate the given equation twice with respect to time.
Step 1: First differentiation
x = 3t^3 + 2t^2 + 5t - 9
dx/dt = 9t^2 + 4t + 5
Step 2: Second differentiation
d^2x/dt^2 = 18t + 4
Step 3: Substituting the value of t as 5 in the above equation, we get:
d^2x/dt^2 = 18(5) + 4 = 94 cm/s^2
Therefore, the acceleration of the particle at the end of 5 seconds is 94 cm/s^2.
Conclusion:
The particle's acceleration is the second derivative of the distance equation with respect to time. By substituting the value of time t as 5 seconds, we can find the acceleration of the particle at the end of 5 seconds. This is an important concept in physics and is used to calculate the motion of objects in different scenarios.
To find the acceleration of the particle after 5 seconds, we need to differentiate the given equation twice with respect to time.
Step 1: First differentiation
x = 3t^3 + 2t^2 + 5t - 9
dx/dt = 9t^2 + 4t + 5
Step 2: Second differentiation
d^2x/dt^2 = 18t + 4
Step 3: Substituting the value of t as 5 in the above equation, we get:
d^2x/dt^2 = 18(5) + 4 = 94 cm/s^2
Therefore, the acceleration of the particle at the end of 5 seconds is 94 cm/s^2.
Conclusion:
The particle's acceleration is the second derivative of the distance equation with respect to time. By substituting the value of time t as 5 seconds, we can find the acceleration of the particle at the end of 5 seconds. This is an important concept in physics and is used to calculate the motion of objects in different scenarios.
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A particle moves in such a way that its distance from a fixed point,at...
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A particle moves in such a way that its distance from a fixed point,at any instant "t," is given by: x=3t^3 2t^2 5t-9 where "x'" is in centimeters and "t'" in second.What will be its acceleration at the end of "5" second:?
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A particle moves in such a way that its distance from a fixed point,at any instant "t," is given by: x=3t^3 2t^2 5t-9 where "x'" is in centimeters and "t'" in second.What will be its acceleration at the end of "5" second:? 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 in such a way that its distance from a fixed point,at any instant "t," is given by: x=3t^3 2t^2 5t-9 where "x'" is in centimeters and "t'" in second.What will be its acceleration at the end of "5" second:? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in such a way that its distance from a fixed point,at any instant "t," is given by: x=3t^3 2t^2 5t-9 where "x'" is in centimeters and "t'" in second.What will be its acceleration at the end of "5" second:?.
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