Question

5) The position of a particle is given by x = 4.50 e-0.30t, where x has units of meters. What is (a) the average velocity between 2.00 and 3.00 s; (b) the instantancous velocity at 2.00 s; (c) the instantancous acceleration at 2.00 s.

Answer 1

The position is given as,

$x(t) = 4.50 e^{-0.30t}$

(a) average velocity can be calculated by finding the positions at given time, this can be done as,

$v_{avg} = \frac{x(3) - x(2)}{3-2} = 4.50e^{-0.30 \times 3} - 4.50 e^{-0.30 \times 2} = -0.64 \, m/s$

(b)(c) The instantaneous velocities and acceleration can be calculated by differentiating and double differentiating the above equations,

$v(t) = \frac{\mathrm{d} }{\mathrm{d} t}x(t) \Rightarrow v(t) = \frac{\mathrm{d} }{\mathrm{d} t} (4.50e^{-0.30t}) = -1.35e^{-0.30 t}$

$a(t) = \frac{\mathrm{d} }{\mathrm{d} t}v(t) \Rightarrow v(t) = \frac{\mathrm{d} }{\mathrm{d} t} (-1.35e^{-0.30t}) = 0.405e^{-0.30 t}$

Therefore the instantaneous velocity and acceleration at t=2s is,

$\Rightarrow v(2) = -1.35e^{-0.30 \times 2} = -0.74 \, m/s$

$\Rightarrow a(2) = 0.405e^{-0.30 \times 2} = 0.22\, m/s^2$