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A particle moves along a curve whose parametric equations are : x = t3+2t,y = −3e−2t and z = 2sin(5t), where x,y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is cm2/s
- a)12 cm2/s
- b)11 cm2/s
- c)10 cm2/s
- d)9 cm2/s
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A particle moves along a curve whose parametric equations are : x = t...
Correct answer is 12 Given equation of motion in 3 different direction
i.e x = t3+2t,y = −3e−2t and z = 2sin5t(cm)
So Velocity in x,y,z directions are Vx = ∂x/∂t = (3t2 + 2)i;Vy = (6e−2t)j;Vz = (10cos5t)kcm/s
Now accelerations in x, y , z directions are
Most Upvoted Answer
A particle moves along a curve whose parametric equations are : x = t...
Given parametric equations:
x = t^3 - 2t
y = -3e^(-2t)
z = 2sin(5t)
To find the magnitude of acceleration at t = 0, we need to find the acceleration vector and then calculate its magnitude.
Acceleration vector:
The acceleration vector can be obtained by taking the second derivative of the position vector with respect to time.
Position vector:
r(t) = (x(t), y(t), z(t)) = (t^3 - 2t, -3e^(-2t), 2sin(5t))
First derivative:
v(t) = (x'(t), y'(t), z'(t)) = (3t^2 - 2, 6e^(-2t), 10cos(5t))
Second derivative:
a(t) = (x''(t), y''(t), z''(t)) = (6t, -12e^(-2t), -50sin(5t))
Calculating acceleration at t = 0:
Substitute t = 0 into the acceleration vector to find the acceleration at t = 0.
a(0) = (6(0), -12e^(-2(0)), -50sin(5(0)))
= (0, -12, 0)
Magnitude of acceleration:
The magnitude of a vector can be found using the formula:
|a| = sqrt(a_x^2 + a_y^2 + a_z^2)
Substituting the values, we get:
|a(0)| = sqrt(0^2 + (-12)^2 + 0^2)
= sqrt(0 + 144 + 0)
= sqrt(144)
= 12
Therefore, the magnitude of the acceleration of the particle at t = 0 is 12 cm/s^2.
Hence, the correct answer is option 'A' (12 cm^2/s).
x = t^3 - 2t
y = -3e^(-2t)
z = 2sin(5t)
To find the magnitude of acceleration at t = 0, we need to find the acceleration vector and then calculate its magnitude.
Acceleration vector:
The acceleration vector can be obtained by taking the second derivative of the position vector with respect to time.
Position vector:
r(t) = (x(t), y(t), z(t)) = (t^3 - 2t, -3e^(-2t), 2sin(5t))
First derivative:
v(t) = (x'(t), y'(t), z'(t)) = (3t^2 - 2, 6e^(-2t), 10cos(5t))
Second derivative:
a(t) = (x''(t), y''(t), z''(t)) = (6t, -12e^(-2t), -50sin(5t))
Calculating acceleration at t = 0:
Substitute t = 0 into the acceleration vector to find the acceleration at t = 0.
a(0) = (6(0), -12e^(-2(0)), -50sin(5(0)))
= (0, -12, 0)
Magnitude of acceleration:
The magnitude of a vector can be found using the formula:
|a| = sqrt(a_x^2 + a_y^2 + a_z^2)
Substituting the values, we get:
|a(0)| = sqrt(0^2 + (-12)^2 + 0^2)
= sqrt(0 + 144 + 0)
= sqrt(144)
= 12
Therefore, the magnitude of the acceleration of the particle at t = 0 is 12 cm/s^2.
Hence, the correct answer is option 'A' (12 cm^2/s).
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A particle moves along a curve whose parametric equations are : x = t3+2t,y = −3e−2t and z = 2sin(5t), where x,y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is cm2/sa)12 cm2/sb)11 cm2/sc)10 cm2/sd)9 cm2/sCorrect answer is option 'A'. Can you explain this answer?
Question Description
A particle moves along a curve whose parametric equations are : x = t3+2t,y = −3e−2t and z = 2sin(5t), where x,y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is cm2/sa)12 cm2/sb)11 cm2/sc)10 cm2/sd)9 cm2/sCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A particle moves along a curve whose parametric equations are : x = t3+2t,y = −3e−2t and z = 2sin(5t), where x,y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is cm2/sa)12 cm2/sb)11 cm2/sc)10 cm2/sd)9 cm2/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves along a curve whose parametric equations are : x = t3+2t,y = −3e−2t and z = 2sin(5t), where x,y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is cm2/sa)12 cm2/sb)11 cm2/sc)10 cm2/sd)9 cm2/sCorrect answer is option 'A'. Can you explain this answer?.
A particle moves along a curve whose parametric equations are : x = t3+2t,y = −3e−2t and z = 2sin(5t), where x,y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is cm2/sa)12 cm2/sb)11 cm2/sc)10 cm2/sd)9 cm2/sCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A particle moves along a curve whose parametric equations are : x = t3+2t,y = −3e−2t and z = 2sin(5t), where x,y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is cm2/sa)12 cm2/sb)11 cm2/sc)10 cm2/sd)9 cm2/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves along a curve whose parametric equations are : x = t3+2t,y = −3e−2t and z = 2sin(5t), where x,y and z show variations of the distance covered by the particle (in cm) with time t (in s). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is cm2/sa)12 cm2/sb)11 cm2/sc)10 cm2/sd)9 cm2/sCorrect answer is option 'A'. Can you explain this answer?.
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