Question

The body speed is given by the equation v (t) = 3x2 - 4x + 5. Then the path traversed by the body in 2 s from the beginning of the movement is ...

Answer 1

Answer:

Acceleration is the rate of change (with respect to time) in velocity. And velocity is the rate of change (with respect to time) in dispalcement.

So at any given instant, if you know the particle’s displacement function x(t), you can take its derivative at that instant dx/dt = v(t) to find the particle’s instantaneous velocity. Similarly, you can take the 2nd derivative ((d^2)x)/(dt^2) = dv/dt = a(t) to find its acceleration in that instant, t.

The question gives us:

v(x) = dx/dt = x^3 - 4x^2 + 6x

Therefore a = dv/dt = (d/dt)(x^3 - 4x^2 +6x)

a = 3(x^2)(dx/dt) - 8x(dx/dt) + 6(dx/dt)

a = 3(x^2)(v(x)) - 8x(v(x)) + 6v(x)

a = (3(x^2) - 8x + 6)(v(x))

Substitute our expression for v(x) into our expression for a. Then substitute x = 2 into the resulting expression and you can calculate the instantaneous acceleration, which is the answer.