a(i). Since you are given a velocity v. time graph, the distance will be represented by:
[tex]\vec{s}(t) = \int_{a}^{b}\left \| \vec{v}(t) \right \|[/tex]
In this case, however, we can just use simple geometry to evaluate the area under the graph v(t). I split it up into 2 trapezoids, and 1 rectangle. So, the area will be as follows:
[tex]A_t = A_{t1}+A_{t2}+A_r[/tex]
[tex]A_{t1} = \frac{30+15}{2}(10) = 225m [/tex]
[tex]A_{t2} = \frac{15+25}{2}(15) = 300m[/tex]
[tex]A_r = 25(30) = 750m[/tex]
[tex]A_t = 750+300+225 = 1275m[/tex]
So, the particle traveled a total of 1275m assuming it never turned back (because it says to calculate distance).
a(iii). Deceleration is a word for negative acceleration. Acceleration is the first derivative of velocity, and so deceleration is too. So, we just need to find the slope of the line that passes through t = 30 because it has a linear slope (meaning the slope doesn't change). So, we can just use simple algebra instead of calculus to figure this out. Recall from algebra that slope (m):
[tex]m = \frac{y_2-y_1}{x_2-x_1} [/tex]
So, let's just pick values. I'm going to pick (25, 30) and (35, 15). Let's plug and chug:
[tex]m = \frac{15-30}{35-25} = -\frac{15}{10} = -\frac{3}{2} [/tex]
Since it's a negative value, this means that acceleration is negative but deceleration is positive (because deceleration is negative acceleration). So, your answer is: The deceleration of the particle at t = 30s is 3/2 or 1.5.
