The displacement ''x'' and time of travel ''t'' for a particle moving on a straight line are related as t=sqare root (x+1)(x-1). Its acceleration at time ''t'' is?Please show the solution.a)1/x-1/xsquareb)1/xcubec)-tsquare/xcubed)-t/xsquare

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: A 4 kg ball on a string is rotated about a circle of radius 10 m. The maximum tension allowed in the string is 52 N. What is the maximum speed of the ball?
Solution:

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=>  t^2=x^2-1

=> x=sqr rt t^2+1

=> differentiate it twice with respect to time to get the accelaration

=> the ans will be  [(t^2+1) - t^2] / [(t^2+1)(sqr rt t^2+1)]

=> now substitute t^2 = x^2-1 to get acc = 1/x^3

Approved