(a)
The displacement of the particle t = 0 s is zero because the particle remain at the position 0.080 m, therefore d₀ = 0 m.
The displacement of the particle by the time t = 1s is:
d₁ = x₂ – x₁
Substitute 0.050 m for x₂ and 0.080 m for x₁,

d₁ = 0.050 m- 0.080 m
= -0.030 m






The displacement of the particle by the time t = 5 s is:
d₅ = x₂ – x₁
Substitute 0.13 m for x₂ and 0.080 m for x₁,
D₅ = 0.13 m -0.080 m
Therefore, the particle displaces by another 0.050 m, relative to d₄.


It can be seen from the figure above that the particle first displaces towards the negative y axis, and then moves towards the positive y axis.





(c)
The slope of the curve at point in displacement time graph is calculated as the tangent of the angle subtended by the tangent to the point, measured relative to the horizontal axis. It is important to note that the counter-clockwise measurement is taken to be as positive, while the clockwise measurement is taken to be negativ
(d)
The slope has the units of velocity, and the plot of slope of the displacement time graph against the time is shown in the graph below:


It is important to note that the graph is a straight line.

(e)
The acceleration of the particle is the slope of the velocity time graph made in part (d), and since the velocity time graph is a straight line, its slope will have a constant value, given as the change of velocity over the change of time.
Therefore, the slope of velocity time graph in part (d) is: