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.