The kinematic equation for the coordinate x(t) of a moving particle is given in the following implicit form: sin(x) = t 2 + t

Question

The kinematic equation for the coordinate x(t) of a moving particle is given in the following implicit form: sin(x) = t 2 + t

Find all possible values for the velocity vx and the acceleration ax of the particle at the moment t = 0 (if we know that as t = 0 the acceleration is positive).

Answer 1

sin(x) = t² + t

differentiate implicitly:

cos(x) dx/dt = 2t + 1

dx/dt = (2t+1)/cos(x)

Make a right triangle with vertical side t²+t and hypotenuse 1 (That means sin(x) = t²+t). The cos(x) will be sqrt(1-(t²+1)²)/1 = sqrt(1-(t²+1)²) having used the definition of the cos and that the horizontal side from Pyth. Thm.

dx/dt = (2t+1)/sqrt(1-(t²+t)²) Plug in t = 0 to obtain v(0) = +/- 1 (we may have to eliminate one of these in order to satisfy a > 1)

You could take the derivative of this to calculate d²x/dt² using quotient rule or you can use the form with cos(x) in it, except make it multiplication by sec(x) rather than division by cos(x) and use product rule:

d(dx/dt)/dt = d/dt ((2t+1)sec(x)) = (2sec(x) + (2t+1)sec(x)tan(x))dx/dt

Maybe easiest is just to implicitly take the derivative of

cos(x)dx/dt = 2t + 1

-sin(x)dx/dt +cos(x)d²x/dt² = 2

Substitute for dx/dt, cos(x), and sin(x) and solve for d²x/dt² as a function of t.

Pick your poison and sub in t = 0 to find a(0) and only take the positive value. Plot the curve in Desmos to see that there are an infinite number of x values, but only 2 possible v and a values.