relation between time "t" and distance "x' is t=ax^2+bx where a and b are constant then acceleration is A / -2abv^2 b/ 2bv^3 C // -2av^3 D // 2av^2
To find the acceleration, we need to differentiate the given equation with respect to time.
Let's differentiate the equation t = ax^2 + bx with respect to time (t):
d/dt(t) = d/dt(ax^2 + bx)
The derivative of t with respect to t is 1, so the left side becomes:
1 = d/dt(ax^2 + bx)
Next, we apply the chain rule for differentiation. The chain rule states that if y = f(u), and u = g(x), then dy/dx = (df/du)(du/dx).
In our equation, t is the dependent variable, and x is the independent variable. So, we can perform the differentiation as follows:
1 = (d/dx(ax^2 + bx))(dx/dt)
Now, let's differentiate ax^2 + bx with respect to x:
d/dx(ax^2 + bx) = 2ax + b
And dx/dt represents the rate of change of x with respect to time. In other words, dx/dt is the speed, or velocity (v). So, we can rewrite the equation as:
1 = (2ax + b)(dx/dt)
Since dx/dt represents velocity (v), we can substitute it into the equation:
1 = (2ax + b)v
Finally, we can rearrange the equation to solve for acceleration (a):
a = (1/(2x)) - (b/(2xv))
So, the answer would be a/(-2bv). Therefore, the correct option is A) -2abv^2.