Answer to Question #183712 in Physics for Hangal

Question #183712

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.



1
Expert's answer
2021-04-22T10:54:46-0400

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\\\\\nv(2.5s) = 4\\cdot 2.5s - 4 = 6"

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


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