Answer
(a) Yes, the particle has a turning point at t = 1 s.
(b) At t = 2 s, $x = 0~m$
At t = 4 s, $x = 16~m$
Note that the initial position of the particle is 10 m, therefor the final answers are:
At 2 s, $x = 10~m$ and,
At 4 s, $x = 26~m$
Work Step by Step
(a) Yes, the particle has a turning point at t = 1 s.
Note that at t = 1 s, the velocity is zero as it changes from a negative velocity to a positive velocity. This shows that the particle has a turning point at this time.
(b) The area between the velocity versus time graph and the x-axis is equal to the displacement.
The area under the curve from t = 0 to t = 2 is -2 + 2, which is 0
From t = 0 to t = 2 s:
$\Delta x = 0$
Therefore, at t = 2 s:
$x = 0~m$
The area under the curve is -2 + the area under t = 1 to t = 3
Therefore, at t = 4 s:
$x = 0.5(3*12)-2~m$
$x = 16~m$
