Question

This is an ADDMATHS question!! It's a kinematics problem. Please help me for both a(i) and a(ii).

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

Answer 1

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

`\vec{s}(t) = \int_(a)^(b)\left \| \vec{v}(t) \right \|`

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:

`A_t = A_(t1)+A_(t2)+A_r`

`A_(t1) = (30+15)/(2)(10) = 225m`

`A_(t2) = (15+25)/(2)(15) = 300m`

`A_r = 25(30) = 750m`

`A_t = 750+300+225 = 1275m`

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):

`m = (y_2-y_1)/(x_2-x_1)`

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

`m = (15-30)/(35-25) = -(15)/(10) = -(3)/(2)`

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.