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The relation between time t and distance x is t=ax
v^2 =1/(2ax+b)^2.put upper equation .you can get answer -2a v^3
ax2+bx=t
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!
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.
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
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.
2a
plz approve..!!
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.
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