The displacement (in meters) of a particle moving in a straight line is given by the equation of motion below,

Question

s=4t^3 where t is measured in seconds. Find the average velocity of the particle over the time interval [6,8].
Find the (instantaneous) velocity of the particle when t = 6.

Answer 1

For the average velocity over [6,8], compute (s(8) - s(6))/2. For the instantaneous velocity at t = 6, find the derivative ds/dt, and evaluate the derivative at t = 6.

Answer 2

Forgetting matters that would desire to do with waves for a sec and thinking approximately this as basically being a component in area (im telling us the two this:) ), the challenge seems some thing like this: Shortest distance to the particle is 6/t^2 . At t=a million its 6 meters away, t=2, a million.5 meters away, and so on. So its getting nearer to us. you will detect the linked value of a place by potential of utilising taking the by potential of-made up of it. Thats going to look some thing like -12/t^3. Its getting nearer and its slowing down. the linked value at t=a is -12/a^3. basically take the by potential of-product and plug on your t's. Thats all I even have been given :). stable luck!