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The relation between time t and distance x is t=ax ^2+bx where a and b are constants.then the acceleration will be ??

The relation between time t and distance x is t=ax

^2+bx where a and b are constants.then the acceleration will be  ??


9 Answers

jaymin prajapati
31 Points
7 years ago

if you differesiate 1=2ax dx/dt+b dx/dt. then you common dx/ your equation will be made1/(2ax+b) =v.then again differensiate.A = 2av(-1)/(2ax+b)^2

                                      v^2 =1/(2ax+b)^2.put upper equation .you can get answer -2a v^3

Vikas TU
14149 Points
7 years ago



Acceleration is double differntiation or derivative for the relation displacement and time given:

thus, dx/dt =>  (2ax + b )dx = dt

now dv/dt = 2a thus, acc. is ''2a''


plz approve!

37 Points
7 years ago

 Just do dx/dt , you will get v then repeat the step by doing dv/dt , you will get acceleration.


dx/dt = 2ax + b

dv/dt = 2a + 0 = 2a

that''s the answer.

Utkarsh Verma
33 Points
7 years ago

A = dv/dt = d2x/dt2

where A is the acceleration.

By finding the second derivative of the given expression, we will find the acceleration.

t = ax2 + bx

1 = 2ax dx/dt + b dx/dt

dx/dt = 1/(2ax+b)

d2x/dt2 = -2a/(2ax+b)2



A = d2x/dt2 = -2a/(2ax+b)2


Prajwal kr
49 Points
7 years ago

Differentiating with respect to time:

1=2ax(dx/dt) + b(dx/dt)

=> v=1/(2ax + b)

=>a=0.(Differentiating wrt time, no t in RHS)


Therfore, a=0.

vivek kumar
57 Points
7 years ago


vishal ojha
35 Points
7 years ago

dx/dt = 2ax + b

dv/dt = 2a + 0 = 2a

plz approve..!!

37 Points
7 years ago

the given expression is a quadratic in x.

to get the acceleration we need to find x as a function of t.

solving the quadratic gives x = [{b2 + 4at}1/2 - b]/2a . 

differentiating twice with respect to t we get the acceleration as -2a*{b2 + 4at}-3/2.

19 Points
3 years ago
dt/dx = 2ax +b dx/dt = 1/2ax+bV=1/2ax+b. .....(I)again differentiate,A=2av(-1)/(2ax+b)^2Put value of v from. eq 1 Final answer : -2av^3

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