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The distance travelled by an object along the axes are given by x = 2t2, y = t2 - 4t, z = 3t - 5. The initial velocity of the particle is
- a)10 unit
- b)12 unit
- c)5 unit
- d)2 unit
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The distance travelled by an object along the axes are given by x = 2t...
Given data:
x = 2t^2
y = t^2 - 4t
z = 3t - 5
To find:
Initial velocity of the particle
Solution:
The velocity of the particle can be found by differentiating the given equations of motion with respect to time.
x = 2t^2
dx/dt = 4t
y = t^2 - 4t
dy/dt = 2t - 4
z = 3t - 5
dz/dt = 3
Therefore, the velocity vector of the particle at any given time t is given as:
v = (4t)i + (2t - 4)j + 3k
To find the initial velocity, we can substitute t = 0 in the above equation.
v(0) = (4 * 0)i + (2 * 0 - 4)j + 3k
v(0) = -4j + 3k
The magnitude of the initial velocity vector is given by:
|v(0)| = √((-4)^2 + 3^2) = √(16 + 9) = √25 = 5
Therefore, the correct option is (c) 5 unit.
x = 2t^2
y = t^2 - 4t
z = 3t - 5
To find:
Initial velocity of the particle
Solution:
The velocity of the particle can be found by differentiating the given equations of motion with respect to time.
x = 2t^2
dx/dt = 4t
y = t^2 - 4t
dy/dt = 2t - 4
z = 3t - 5
dz/dt = 3
Therefore, the velocity vector of the particle at any given time t is given as:
v = (4t)i + (2t - 4)j + 3k
To find the initial velocity, we can substitute t = 0 in the above equation.
v(0) = (4 * 0)i + (2 * 0 - 4)j + 3k
v(0) = -4j + 3k
The magnitude of the initial velocity vector is given by:
|v(0)| = √((-4)^2 + 3^2) = √(16 + 9) = √25 = 5
Therefore, the correct option is (c) 5 unit.
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The distance travelled by an object along the axes are given by x = 2t2, y = t2 - 4t, z = 3t - 5. The initial velocity of the particle isa)10 unitb)12 unitc)5 unitd)2 unitCorrect answer is option 'C'. Can you explain this answer?
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