I've calculated the displacements and average velocities of both particles.

$A$ average velocity $= 1.1180m/s$ displacement is $11.1803m$.
$B$ average velocity $= 0m/s$ displacement is $0m$.

When, approximately, are the particles at the same position?

I can't figure out how to find the when they are at the same position. I can look at the graph and tell it's around the $1.75s$ mark.

When, approximately, do the particles have the same velocity? 

It's when both angles are the same, around $6s$.

Graph of the particles.


I agree with Michael that the question is probably expecting you to estimate the answers from the graph. However, this is what I would do if these were experimental results I was analysing.

Let's assume that the blue line is a straight line $f(x)$ and the curve is a quadratic $g(x)$. A straight line has the form $y = ax + b$ where $a$ is the gradient and $b$ is the $y$ intercept. From the graph the $y$ intercept is 1 and the gradient is 4/10 i.e. 0.4. Then:

$$ f(x) = 0.4x + 1 $$

The quadratic is a little harder, but if you know the two zeroes of the quadratic, $x_1$ and $x_2$ then the function has the form $g(x) = A(x - x_1)(x - x_2)$ where $A$ is some constant. In our case the two zeros are both $x = 5$ so the function is $g(x) = A(x - 5)^2$. To find the constant $A$ note that when $x = 0$ $y = 4$ so the constant $A$ must be 4/25 or 0.16.

$$ g(x) = 0.16(x - 5)^2 = 0.16x^2 - 1.6x + 4 $$

So to find the two values of $x$ when the curves cross we just set $f(x) = g(x)$:

$$ 0.4x + 1 = 0.16x^2 - 1.6x + 4 $$

and a quick rearrangement gives:

$$ 0.16x^2 - 2x + 3 = 0 $$

To get the two solutions to this use the quadratic formula and you find the curves cross at $x \approx 1.743$ and $x \approx 10.757$.

As you say, the two particles have the same velocity when the gradients are the same i.e. $f^'(x) = g'(x)$. Differentiating our expressions for $f(x)$ and $g(x)$ gives:

$$ f^'(x) = 0.4 $$

$$ g^'(x) = 0.32x - 1.6 $$

Set these equal to find the point where the slopes are equal:

$$ 0.4 = 0.32x - 1.6 $$


$$ x \approx 6.25 $$

Incidentally, I'm a bit concerned by your calculation of the average velocity. Unless there's a bit to the question you haven't posted, the average velocity of A is distance moved (4 metres) divided by time take (10 secs) so the average velocity is 0.4 m/sec not 1.118.

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I believe the questions are asking you "approximately" for a reason, you are supposed to estimate this from the graphs. If you had an actual equation which creates the graph for B, this would be simple. (A is easy to get using point-slope).

Same position is simply the two equations added/subtracted from each other to solve for when x is the same in both.

The velocity is, as you mentioned, whenever the slope is the same. You calculate this by setting the derivatives of the two equations equal to each other and seeing which value of x results.

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