At time t=0 a particle starts travelling from a height 7 z cm in a plane keeping z coordinate constant. At any instant of time it's position along the x and y directions are defined as 3 t and 5 t³ respectively. At t = 1s acceleration of the particle will be

Solution

Physics sir · 4 steps

Differentiate each coordinate twice; any coordinate that is constant or linear in t drops out.

1. Write the position. The particle has x = 3t, y = 5t³ and z = 7 (constant), all in cm. So r⃗ = 3t x + 5t³ y + 7 z.
2. First derivative. Velocity is v⃗ = dr⃗/dt = 3 x + 15t² y. The constant z gives zero.

   a⃗ = dv⃗/dt = 0 x + 30t y

3. Second derivative. The x velocity is a constant 3, so aₓ = 0. Only the y part survives: a_y = 30t.
4. So. At t = 1 s, a⃗ = 30 y cm s⁻², which is option (b).

The idea to keep. Acceleration needs two derivatives of position. A constant coordinate vanishes after one, a linear one after two, so only powers of t higher than 1 survive.

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