A particle moves in a straight line so that its displacement x at any time t is given by x²=1+t². Its acceleration at any time t is x⁻ⁿ where n= _________.

Solution

Physics sir · 4 steps

Do not solve for x first: differentiate x² = 1 + t² as it stands, twice, and let x² - t² = 1 tidy up.

1. First derivative. Differentiate both sides with respect to t: 2xdx/dt = 2t. With v = dx/dt, this gives xv = t, so v = t/x.
2. Second derivative. Differentiate xv = t once more, using the product rule. The derivative of x is v and the derivative of v is a, so v · v + x · a = 1.

   a = (1 - v²)/x = (1 - t²/x²)/x = (x² - t²)/x³

3. Use the given relation. From x² = 1 + t², the numerator is x² - t² = 1. So a = 1/x³ = x⁻³.
4. So. Comparing a = x⁻³ with x⁻ⁿ gives n = 3.

The common slip: 1. Treating x as a constant while differentiating xv = t drops the v² term, giving a = 1/x and n = 1. Both x and v change with time, so use the product rule.

The idea to keep. When position is given as an implicit relation, differentiate the relation itself with respect to time. Then use the original relation to clean up the result.

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