This is an addmaths it's a kinematics problem. me for both a(i) and a(ii).

i think i know a(i) but i need to make sure. in advance!

a(i). Since you are given a velocity v. time graph, the distance will be represented by:


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:






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):


So, let's just pick values. I'm going to pick (25, 30) and (35, 15). Let's plug and chug:


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.

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