The displacement (in meters) of a particle moving in a straight line is given by the equation of motion:

s = 4/t^2, where t is measured in seconds.

Required:
Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.

Respuesta :

Answer:

  • At [tex]t = 1\; \rm s[/tex], the particle should have a velocity of [tex]-8\; \rm m \cdot s^{-1}[/tex].
  • At [tex]t = 2\; \rm s[/tex], the particle should have a velocity of [tex]-1\; \rm m \cdot s^{-1}[/tex].
  • At [tex]t = 3\; \rm s[/tex], the particle should have a velocity of [tex]\displaystyle -\frac{8}{27}\; \rm m \cdot s^{-1}[/tex].

For [tex]a > 0[/tex], at [tex]t = a \; \text{second}[/tex], the particle should have a velocity of [tex]\displaystyle -\frac{8}{a^3}\; \rm m \cdot s^{-1}[/tex].

Explanation:

Differentiate the displacement of an object (with respect to time) to find the object's velocity.

Note that the in this question, the expression for displacement is undefined (and not differentiable) when [tex]t[/tex] is equal to zero. For [tex]t > 0[/tex]:

[tex]\begin{aligned}v &= \frac{\rm d}{{\rm d}t}\, [s] = \frac{\rm d}{{\rm d}t}\, \left[\frac{4}{t^2}\right] \\ &= \frac{\rm d}{{\rm d}t}\, \left[4\, t^{-2}\right] = 4\, \left((-2)\, t^{-3}\right) = -8\, t^{-3} =-\frac{8}{t^3}\end{aligned}[/tex].

This expression can then be evaluated at [tex]t = 1[/tex], [tex]t = 2[/tex], and [tex]t = 3[/tex] to obtain the required results.