The velocity of a particle moving along the x-axis varies in
time according to the expression vx(t) = (40 – 5 t2) m/s, where t is in
seconds.
Find the average acceleration in the time interval t=0 to t=2.0s.
Determine the acceleration at t = 2.0 s.
1
Expert's answer
2013-01-29T08:24:05-0500
Acceleration is the rate at which the velocity of a body changes with time: a = dv/dt a - acceleration v - velocity In our case: ax = dvx(t)/dt ax = d/dt (40 - 5 t2) = -10t the average acceleration in the time interval t=0 to t=2.0s: a_av = [ax(0) + ax(2)]/2 = (0-20)/2 = -10m/s^2 the acceleration at t = 2.0 s ax(2) = -10*2 = -20 m/s^2
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments
Dear visitor, please use panel for submitting new questions
What is the particle position at 2s
Dear Eman, it is the properties of derivation: (40 - 5*t^2)' = (0 + 2*5*t) = 10t
Why the ax= -10 and not -20 Please explain
Leave a comment