Position of an ant (S in metres) moving in Y-Z plane is given by S=2 t² ĵ+5 k̂ (where t is in second). The magnitude and direction of velocity of the ant at t=1 s will be :

Solution

Physics sir · 3 steps

Differentiate each component of position; the constant 5 k̂ term drops out completely.

1. Differentiate. Velocity is v⃗ = dS⃗/dt. Differentiate each component separately: 2t² gives 4t, and the constant 5 gives 0.

   v⃗ = d/dt(2t² ĵ + 5 k̂) = 4t ĵ

2. Substitute the time. At t = 1 s, v⃗ = 4 × 1 ĵ = 4 ĵ m s⁻¹. Only the ĵ component survives, so the ant moves along the y-axis.
3. So. The velocity is 4 m s⁻¹ in the y-direction, which is option (d).

The idea to keep. Differentiate a position vector component by component. A constant component, like a fixed height, only fixes where the body is and adds nothing to its velocity.

Answer from the NTA final answer key. Solution by PrepWiser.