Answer to Question #183712 in Physics for Hangal

Question

A body moves along the x-axis. Its position is expressed in the given equation with respect to time: x = -4t + 2t2

. With the given time intervals of t = 0 to t = 1 s and t = 1 s to t = 3 s, find the body’s a) displacement b) average velocity and c) instantaneous velocity at t =

2.5 s.

Answer 1

1. Let's consider the interval $[0s, 1s]$. At the beginning of this interval the position of the object is:

$$x_1 = -4\cdot 0s + 2\cdot (0s)^2 = 0$$

At the end:

$$x_2 = -4\cdot 1s + 2\cdot (1s)^2 = -2$$

Thus, the displacement is:

$$d_1 = x_2 - x_1 = -2$$

By definition, the averabe velocity is:

$$v_1 = \dfrac{d_1}{1s - 0s} = \dfrac{-2}{1s - 0s} = -2$$

2. Let's consider the interval $[1s, 3s]$. At the beginning of this interval the position of the object is $x_2 = -2$ , at the end:

$$x_3 = -4\cdot 3s + 2\cdot (3s)^2 = 6$$

Thus, the displacement is:

$$d_2 = x_3 - x_2 = 8$$

By definition, the averabe velocity is:

$$v_2 = \dfrac{d_2}{3s - 1s} = \dfrac{8}{3s - 1s} =4$$

3. The instantaneous velocity is defined as the derivative of position with respect to time. Thus:

$$v(t) = \dfrac{dx}{dt} = 4t-4\\ v(2.5s) = 4\cdot 2.5s - 4 = 6$$

Answer. a) Displacements: -2 and 8, b) average velocities: -2 and 4, c) instantaneous velocity: 6.