A particle starts from origin at t=0 with a velocity 5 î m / s and moves in x-y plane under action of a force which produces a constant acceleration of (3 î+2 ĵ) m / s². If the x-coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is √α m / s. The value of α is _________.

Solution

Physics sir · 4 steps

Use the x-motion to find the time first; then build both velocity components and combine them.

1. Find the time. Along x, uₓ = 5 m s⁻¹ and aₓ = 3 m s⁻². Set x = 84 m in x = uₓ t + 1/2aₓ t².

   84 = 5t + 3/2t² ⇒ 3t² + 10t - 168 = 0 ⇒ t = (-10 + 46)/6 = 6 s

2. Velocity components. vₓ = uₓ + aₓ t = 5 + 3 × 6 = 23 m s⁻¹. Along y the particle starts from rest, so v_y = a_y t = 2 × 6 = 12 m s⁻¹.
3. Combine them. Speed = √(vₓ² + v_y²) = √(23² + 12²) = √(529 + 144) = √673 m s⁻¹.
4. So. α = 673.

The common slip: 529. Forgetting the y-component of velocity and using only vₓ = 23 m s⁻¹ gives 529, not 673.

The idea to keep. With constant acceleration in a plane, treat x and y as two separate straight-line motions sharing the same time t. Find t from one direction, then use it in the other.

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