The position vector of a particle is r⃗ = 2t2 x̂ + 3t ŷ + 4 ẑ. Then the instantaneous velocity v⃗ and acceleration a⃗ respectively lie
on xy-plane and along z-direction
on yz-plane and along x-direction
on yz-plane and along y-direction
on xy-plane and along x-direction
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CDS-2 – 2018
The instantaneous velocity vector v⃗ is the first derivative of the position vector with respect to time:
v⃗ = dr⃗/dt = d/dt (2t² x̂ + 3t ŷ + 4 ẑ) = (d/dt 2t²) x̂ + (d/dt 3t) ŷ + (d/dt 4) ẑ = 4t x̂ + 3 ŷ + 0 ẑ = 4t x̂ + 3 ŷ.
A vector of the form Ax̂ + Bŷ has components only in the x and y directions. Such a vector lies in the xy-plane.
The instantaneous acceleration vector a⃗ is the first derivative of the velocity vector with respect to time:
a⃗ = dv⃗/dt = d/dt (4t x̂ + 3 ŷ) = (d/dt 4t) x̂ + (d/dt 3) ŷ = 4 x̂ + 0 ŷ = 4 x̂.
A vector of the form Ax̂ has a component only in the x direction. Such a vector lies along the x-direction.
Therefore, the instantaneous velocity v⃗ lies on the xy-plane, and the instantaneous acceleration a⃗ lies along the x-direction.
– Acceleration is the time derivative of the velocity vector.
– A vector is in a plane if it has non-zero components only in the directions defining that plane.
– A vector is along an axis if it has a non-zero component only in the direction of that axis.