Question

What is the instantaneous acceleration at t = 2.1 s of a particle whose position is given by y(t) = -4.1t² + 6.7?

Answer 1

Let's first find the velocity of the particle:

$$v(t)=\dfrac{dy(t)}{dt}=-8.2t.$$

Then, we can find the acceleration of the particle:

$$a(t)=\dfrac{dv(t)}{dt}=-8.2\ \dfrac{m}{s^2}.$$

Since the expression for the acceleration don't depend on $t$ , the acceleration of particle for each $t$

will be $a(t)=-8.2\ \dfrac{m}{s^2}.$ Therefore, the instantaneous acceleration at t = 2.1 s equals $-8.2\ \dfrac{m}{s^2}$.