a particle moves along the x-axis with velocity given by v(t) = 3t^2 + 6t for time t >_. if the particle is at position x = 2 at time t =0, what is the position of the particle at time t = 1?
Velocity is given as a function of time and as you can see it is not linear, so we can’t use formula Distance=Velocity⋅Time.
Instead, we refer to velocity at the point in time, t as the derivative of distance over time, so v(t)=dfracdxdt, where x(t) is the distance covered at time t and t is time.
Integrating this equation we obtain:
v(t) = dx/dt = 3 t^2 + 6t after integrating which implies, x(t)= t^3+3t^2 +c , cis an integrant constant. Now
at t=0, x=2 which gives, c=2, i.e x(t)= t^3+3t^2+2. Hence at t=1, x(t)= 1^3+3*1^2+2=6