Llama N.
asked • 05/15/23What is the velocity and acceleration of an object after 2 seconds if its displacement is the square root of 2t + the cube root of 2t^3?
What is the velocity and acceleration of an object after 2 seconds if its displacement is the square root of 2t + the cube root of 2t^3?
More
1 Expert Answer
Robert C. answered • 05/18/23
Tutor
5.0
(56)
Physics Grad with 3+ years Tutoring experience
,Assuming the displacement s = (2t)^(1/2) + (2t^3)^(1/3), the velocity, v, being the derivative of s, is v=t^(-1/2)+2^(1/3), and a, being the derivative of v, is a = -t^(-3/2)/2 or a=-1/(2*t^(3/2)). you can plug in t=2 to these equations to find the velocity and acceleration of the object at this time.
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
William W.
Do you mean displacement = squareroot(2t) + cuberoot(2t^3) or do you mean displacement = squareroot(2)*t + cuberoot(2)*t^3 or something else?05/16/23