The distance x covered by a particle in one dimensional motion varies with time t as
x² = at² + 2bt + c. If the acceleration of the particle depends on x as x –n , where n is an integer, the value of n is __________

Solution

Physics sir · 4 steps

Differentiate the given relation twice and let the time terms cancel; only a power of x survives.

1. First derivative. Differentiate x² = at² + 2bt + c with respect to t: 2xv = 2at + 2b, so xv = at + b. Call the acceleration α so it does not clash with the constant a.
2. Second derivative. Differentiate xv = at + b once more. The product rule gives v² + xα = a, so α = (a - v²)/x.
3. Remove v. From the first step, v = (at + b)/x. Put this in and write a as (a x²)/x², then use x² = at² + 2bt + c.

   α = (a x² - (at+b)²)/x³ = ((a²t² + 2abt + ac) - (a²t² + 2abt + b²))/x³ = (ac - b²)/x³

4. So. The numerator ac - b² is a constant, so α ∝ x⁻³ and n = 3.

The common slip: 1. Stopping at α = (a - v²)/x and reading the power as 1 ignores that v itself depends on x, which gives 1 instead of 3.

The idea to keep. When the relation is given as x² in terms of t, differentiate it as it stands rather than taking a square root. Keep simplifying until every t cancels out.

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