A small object moves along the x-axis with acceleration ax(t)=−(0.0320m/s3)(15.0s−t). At t=0 the object is at x=−14.0m and has velocity v0x=9.10m/s. What is the x-coordinate of the object when t=10.0s?
Added by Michael A.
Step 1
We can do this by integrating the acceleration function with respect to time. The integral of ax(t) = -(0.0320 m/s^3)(15.0 s - t) dt from 0 to t is: v(t) = ∫ ax(t) dt = ∫ -(0.0320 m/s^3)(15.0 s - t) dt = -(0.0320 m/s^3) * [7.5t^2 - 0.5t^3] + C We know that at Show more…
Show all steps
Your feedback will help us improve your experience
Paul Gabriel and 59 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A small object moves along the x-axis with acceleration ax(t) = -(0.0320 m/s^3)(15.0 s - t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 4.90 m/s. Find x-coordinate of object at t=10.0s
Adi S.
A small object moves along the x-axis with acceleration ax(t) = -(0.0320 m/s^3)(15.0 s - t). At t = 0, the object is at x = -14.0 m and has velocity v0x = 9.10 m/s. (a) What is the x-coordinate of the object when t = 10.0 s? Express your answer with the appropriate units.
Karan S.
Suppose an object moves along a line at $15 \mathrm{m} / \mathrm{s}$ for $0 \leq t<2$ and at $25 \mathrm{m} / \mathrm{s}$ for $2 \leq t \leq 5,$ where $t$ is measured in seconds. Sketch the graph of the velocity function and find the displacement of the object for $0 \leq t \leq 5$
Integration
Approximating Areas under Curves
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD