A particle moves in a straight line and its position x at time t is given by x^{2}=2+t. Its acceleration will be ?????? Share with your friends Share 54 Akansha Tyagi answered this $Position,{x}^{2}=2+t\Rightarrow x=\sqrt{2+t}\phantom{\rule{0ex}{0ex}}Velocity,v=\frac{dx}{dt}=\frac{1}{2}(2+t{)}^{-1/2}\phantom{\rule{0ex}{0ex}}Aceleration,a=\frac{{d}^{2}x}{d{t}^{2}}=\frac{dv}{dt}=\frac{-1}{4}(2+t{)}^{-3/2}m/se{c}^{2}$ 55 View Full Answer