Answer to Question #266719 in Physics for Nishat

Question

A particle moves in a straight line such that its displacement, x meters, from a fixed point O

on the line at time t seconds is given by 𝑥 = 60[𝑒

−2𝑡 − 𝑒

−4𝑡

].

(a) Find the time when the particle is instantaneously at rest.

(b) Find the displacement of the particle from O when t = 4 s.

(c) Find the total distance travelled during the first 4 seconds of its motion.

Answer 1

(a) The time when the particle is resting:

$$v(t)=x'(t)=-120e^{-4t}(e^{2t}-2)=0,\\ t=0.35\text{ s}.$$

(b) Displacement:

$$x(4)=60[e^{−2·4} − e^{−4·4}]=0.02\text{ m}.$$

(c) The total distance:

$$d=\int_0^4\sqrt{1+(v(t))^2}\text dt=31.5\text{ m}.$$