The relation between time t and distance x is t = ax^2 + bx, where a and b are positive constants. Its acceleration is?

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Badiuddin askIITians.ismu Expert · Answer

Dear Vishrant t = ax^2 + bx differentiate w.r.t t 1= 2ax dx/dt + b dx/dt 1= (2ax+b)dx/dt dx/dt = 1/( 2ax+b) again differentiate d 2 x/dt 2 = -1/( 2ax+b) 2 (2adx/dt) a= -1/ ( 2ax+b) 2 * 2a/ ( 2ax+b) a = -2a/ ( 2ax+b) 3 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Askiitians Experts Badiuddin

Samuel Garry · Answer

So, if t = ax^2+bx then, differentiate both side with respect to time (t) dt/dt = 2ax.dx/dt + b.dx/dt 1 = 2axv + bv v.(2ax + b) = 1 (2ax + b) = 1/v Again differentiate both side with respect to time (t) 2a.dx/dt = -v -2 . dv/dt 2av = -v -2 . acceleration acceleration = -2av 3 Hence, retardation= -2av 3

Yash Chourasiya · Answer

Dear Student Please see the solution in the attachment. Given I hope this answer will help you. Thanks & Regards Yash Chourasiya

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The relation between time t and distance x is t = ax^2 + bx, where a and b are positive constants. Its acceleration is?

The relation between time t and distance x is t = ax^2 + bx, where a and b are positive constants. Its acceleration is?

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The relation between time t and distance x is t = ax^2 + bx, where a and b are positive constants. Its acceleration is?

The relation between time t and distance x is t = ax^2 + bx, where a and b are positive constants. Its acceleration is?

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