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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?

Grade:11

3 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

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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Samuel Garry
61 Points
3 years ago
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
Yash Chourasiya
askIITians Faculty 256 Points
11 months ago
Dear Student

Please see the solution in the attachment.
Given
643-907_Untitled.png

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

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