Watch help video: A particle moves along the x-axis with velocity given by v(t) = 3t^2 + 13 for time t ≥ 0. If the particle is at position x = -8 at time t = 1, what is the position of the particle at time t = 2?
Added by Alfonso E.
Step 1
The antiderivative of 3t^2 is t^3, and the antiderivative of 13 is 13t. Therefore, the position function x(t) is given by: x(t) = t^3 + 13t + C where C is the constant of integration. Show more…
Show all steps
Close
Your feedback will help us improve your experience
Sri K and 84 other Calculus 1 / AB 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 particle moves along the x-axis with velocity given by v(t) = 3t^2 - 6t for time t >= 0. If the particle is at position x = -9 at time t = 1, what is the position of the particle at time t = 2?
Sri K.
A particle moves along the x-axis with velocity given by v(t) = 3t^2 - 20 for time t ≥ 0. If the particle is at position x = -7 at time t = 2, what is the position at time t = 3?
Harsha M.
A particle moves along the x-axis so that at time t >= 0 the position of the particle is given by x(t) = 0.5t^4 - 1.5t^3 - 2t^2 + 6t - 1. What is the velocity of the particle at the first instance the particle is at the origin?
Supratim P.
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD