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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?

Vishrant Vasavada , 16 Years ago

Grade 11

3 Answers

Badiuddin askIITians.ismu Expert

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

d²x/dt² = -1/(2ax+b)² (2adx/dt)

a= -1/(2ax+b)²* 2a/(2ax+b)

a = -2a/(2ax+b)³

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Last Activity: 16 Years ago

Samuel Garry

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 = -2av³

Hence, retardation= -2av³

Last Activity: 7 Years ago

Yash Chourasiya

Dear Student

Please see the solution in the attachment.
Given

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

Last Activity: 5 Years ago

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