Velocity Functions & Instantaneous Acceleration Video Lessons

Concept

Problem: A small object moves along the x-axis with acceleration ax(t) = −(0.0320 m/s3)(15.0 s−t). At t = 0 the object is at x = −14.0 m and has velocity exttip{v_{0x}}{v_0x} = 6.70 m/s.What is the x-coordinate of the object when t = 10.0 s?

FREE Expert Solution

We are asked to find the position of an object at a specific time, given an acceleration function and some initial conditions.

Anytime we're given a position, velocity, or acceleration function and asked to find one or more of the others, we know it's a motion problem with calculus. A PVA diagram like the one below can help remind you of the relationships between the three functions:

$\mathbit{P}\begin{array}{c}{\mathbf{←}}\\ {\mathbf{\to }}\end{array}\underset{\frac{\mathbit{d}}{\mathbit{d}\mathbit{t}}}{\overset{{\mathbf{\int }}{\mathbit{d}}{\mathbit{t}}}{\mathbit{V}}}\begin{array}{c}{\mathbf{←}}\\ {\mathbf{\to }}\end{array}\mathbit{A}$

We need to get from an acceleration function to a position function, which means we need to integrate the acceleration twice.

First, we integrate the acceleration function to get to velocity. Make sure to remember the constant of integration, which is going to be the velocity at t = 0 (initial velocity, v0).

Then, to get from velocity to position, we integrate the velocity function. (Again, make sure to add the initial position as a constant of integration!)

We'll also use the power rule of integration:

(for example, $\mathbf{\int }{\mathbit{t}}^{\mathbf{2}}\mathbit{d}\mathbit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}{\mathbit{t}}^{\mathbf{3}}$

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Problem Details

A small object moves along the x-axis with acceleration ax(t) = −(0.0320 m/s3)(15.0 s−t). At t = 0 the object is at x = −14.0 m and has velocity = 6.70 m/s.

What is the x-coordinate of the object when t = 10.0 s?