Differentiation and determine the t of the slops. When is speeding up and when i

Question

Differentiation and determine the t of the slops. When is speeding up and when is slowing down?

I got the answers by gussing but I don't actually understand it. Also, I couldn't find the last answer for the last answer...

At least one of the answers above is NOT correct. 1 of the questions remains unanswered. (1 pt) A particle moving along the -axis, has a location given by a (t) t 20, where t is measured in seconds and at in metres. 1 t The acceleration a(t of the particle is given by the second derivative of its position function a(t). For this particle, a t 2(A2-3voM2+1Y3 t 1.732 The particle is speeding up in the time interval 1 t 1 and t The particle is slowing down for the time intervals 0

Explanation / Answer

x(t)= t/ (1+t2)

a(t) = d2x/dt2= 2t(t2-3)/(t2+1)3

Using a very simple concept. Acceleration is opposite to velocity for slowing down.

For 1<t<1.732 , direction of a(t) is same as v(t)

v(t)= dx/dt = (1-t2)/(1+t2)

For 0<t<1 and t>1.732

Direction of a(t) is opposite to v(t) , so slowing down.

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