Question

A point moves in a straight line so that its displacement x at time t is given by x² = 1 + t². Its acceleration at time t is

Answer 1

Dear Student
$$\begin{gathered} x^2=1\hspace{0.17em}+t^2 \\ Differentiate \\ 2x\frac{dx}{dt}=2t \\ x\frac{dx}{dt}=t \\ \frac{dx}{dt}=\frac{t}{x} \\ Nowtakingdifferentiationforaccelerationweget, \\ x\frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}\frac{dx}{dt}=1 \\ x\frac{d^{2}x}{d{t}^2}+{\left(\frac{dx}{dt}\right)}^2=1 \\ x\frac{d^{2}x}{d{t}^2}+{\left(\frac{t}{x}\right)}^2=1 \\ xa=1-{\left(\frac{t}{x}\right)}^2 \\ a=\frac{1-{\left(\frac{t}{x}\right)}^2}{x}=\frac{x^2-t^2}{x^3} \\ Asgiven,x^2=1\hspace{0.17em}+t^2 \\ a=\frac{1+t^2-t^2}{x^3}=\frac{1}{x^3} \end{gathered}$$
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