The relation between time 't' and distance 'x' is t=α x²+β x, where α and β are constants. The relation between acceleration (a) and velocity (v) is :

Solution

Physics sir · 4 steps

Differentiate t with respect to x to get 1/v, then differentiate once more with respect to time.

1. Get 1/v. Since t is given as a function of x, differentiate with respect to x. This gives dt/dx = 2α x + β, and dt/dx is just 1/v.

   1/v = 2α x + β

2. Differentiate in time. Differentiate both sides with respect to t. The left side gives -1/v²dv/dt = -a/v². The right side gives 2αdx/dt = 2α v, and β drops out.
3. Solve for a. So -a/v² = 2α v. Multiplying both sides by -v² gives a = -2α v³.
4. So. The relation is a = -2α v³, which is option (c).

The idea to keep. When time is given as a function of position, dt/dx = 1/v. Differentiating 1/v in time gives -a/v², so the acceleration appears directly.

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