Question

A particle moves along the x-axis and its position at time t-seconds is found by x(t)=12 + 20t – 2t^2

Find acceleration @ t=6. ( I put -4)

At what time does the particle change direction ? explain.

Find the speed of the particle at t = 6 seconds. is it increasing or decreasing?

Answer 1

The acceleration of the particle is the second derivative of its position function:

`x(t)=12+20t-2t^2\implies x'(t)=20-4t\implies x''(t)=-4`

The particle has constant acceleration and as you said,
`x''(6)=-4`.

The particle changes direction when the sign of its velocity alternates between positive and negative, which means we can find any time this happens by solving for
`x'(t)=0`, then verifying that the sign of
`x'(t)` changes to either side of this value of
`t`.

`x'(t)=20-4t=0\implies t=5`

We have
`x'(4)=4>0` and
`x'(6)=-4<0`, so indeed the velocity changes sign at
`t=5`, so the particle's direction also changes at this point.

We already found that
`x'(6)=-4`, and the speed is the magnitude of velocity, which means that the speed at
`t=6` is 4. The particle has constant acceleration, which means its speed will be increasing.