A particle is moving along the x-axis with its coordinate with the time 't' given be
x(t) = 10 + 8t – 3t². Another particle is moving the y-axis with its coordinate as a function of time given by y(t) = 5 – 8t³.
At t = 1s, the speed of the second particle as measured in the frame of the first particle is given as √v. Then v (in m/s) is ______.

Solution

Physics sir · 4 steps

Find both velocities at t = 1 s, subtract them as vectors, then square the magnitude.

1. First particle. Differentiate x(t) to get vₓ = 8 - 6t. At t = 1 s, v⃗₁ = 2 î m s⁻¹.
2. Second particle. Differentiate y(t) to get v_y = -24t². At t = 1 s, v⃗₂ = -24 ĵ m s⁻¹.

   v⃗₂₁ = v⃗₂ - v⃗₁ = -2 î - 24 ĵ

3. Take the magnitude. The two components are perpendicular, so |v⃗₂₁| = √(2² + 24²) = √(4 + 576) = √580 m s⁻¹.
4. So. Comparing with √v gives v = 580.

The common slip: 676. Adding the two speeds as plain numbers, 2 + 24 = 26, and squaring gives 676, not 580. The velocities are perpendicular, so they combine as √(2² + 24²).

The idea to keep. Velocity of B as seen from A is v⃗_B - v⃗_A. Get each velocity by differentiating, subtract component by component, and only then find the magnitude.

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