I want my graph to be continuous

i want the graph to be continous. However, it is breaking form point 7 to 8 (t=7 to t=8).

as you can see from the function, at first it is a polynomial and after that it is linear, it is not taking 7 as a common point.

How can i fix that ??

.

Answer 1

[Rik]

You can also let Matlab do the heavy lifting by creating an anonymous function and using fplot:

a=8.4;
v=58.8;
fun=@(t) 0 + ...
    (t<=7).*(0.5*a.*(t.^2)) + ...
    (t>7).*(v.*t);
fplot(fun,[0 30])

However, if you want to make the transition smooth, you will have to adjust a or v:

t=7;
[0.5*a.*(t.^2) v.*t]

ans = 1×2

205.8000 411.6000

To make those equal for some value of t, this relation must hold:

0.5*a.*(t.^2)==v.*t
a*t^2=2vt
at=2v
t=2v/a

If you want t=7, that means

7=2v/a ==> v=7a/2 ==> v=29.4

Now lets plot that one instead:

v=29.4;
fun=@(t) 0 + ...
    (t<=7).*(0.5*a.*(t.^2)) + ...
    (t>7).*(v.*t);
fplot(fun,[0 30])

Answer 2

[Wan Ji]

It may work if you do by following Scott MacKenzie's advice or maybe you can do like this

t = [0:1:6, 6.01:0.01:6.99, 7:1:30];

This would make your curve more smooth then.

[Scott MacKenzie]

@Wan Ji You are referring to my first answer which I deleted as I think it was a bit wonky. I posted a new answer which is trivial -- just change 7 to 14 in line 6. Or, there is @Rik's answer which is a bit more complicated but maintains t = 7 as the transition point.

[Wan Ji]

When t = 7, y encounters a gap. So a possible way is just to calculate them both and then to compare them.

a = 8.4; v = 58.8; t=0:0.1:30;
x1 = 0.5*a*t.^2; x2 = v*t;
q = x1<x2;
x = x1.*q + x2.*~q;
plot(t,x)

quite easy to find that the intersection point is 14.2, as mentioned by @Scott MacKenzie.

Answer 3

[Scott MacKenzie]

You need to determine the point where the two functions intersect and then use that point in line 6 as the transition between the two curves. That point is t = 14, which (luckily) happens to be an integer. So, change 7 to 14 in line 6 and you're good to go:

[Rik]

Another possiblity is that the graph should be smooth, meaning both the y-value and the derivative should be equal at both sides of the jump. However, only the trivial solution where everything is 0 satisfies that.

[Bjorn Gustavsson]

@Rik, Or simply continue along the tangent at the transition...

[Rik]

@Bjorn True, in which case v would have to be set at a*t.