Moving dots

dot P: a=2, v=2t+6, s=t^2 +6t (simple integration) dot Q: v=k, s=k(t-2) because of delayed start. Put the s formulas equal which leads to a quadratic. Minimum occurs when discriminant = 0 therefore k = 2 (impossible of course) or k = 18 (ANSWER)

Essentially, distance covered by P = u t + 1/2 a t^2 where u = 6 , a = 2 = 6t + t^2 distance covered by Q = k (t-2)

So at the time of catching up, t^2+6t = k(t-2)

Since we would like to minimalize k, we have to minimalize the ratio (t^2 + 6t ) / (t-2)

Let f(t) = (t^2+6t) / (t-2)

taking derivatives, f'(t) = ((t-2)(2t+6) - (t^2+6t)(1)) / (t-2)^2

Since all we need is the numerator, we set it to 0,

i.e (2t^2-4t + 6t -12) - t^2 - 6t = 0

hence t^2 - 4t -12 = 0

yielding t=6 or t=-2

Since t=-2 is not possible, t = 6 is the only solution.

Now, we substitute back the value of t=6 in t^2+6t = k(t-2) to get

36 + 36 = 4*k

giving k = 72/4 = 18 m/s

The distance as a function of time for dot P, say s(P) = 6t + t^2 (from s = ut + 1/2 at^2)

And as dot Q starts 2 seconds later with a constant velocity, s(Q) = k (t-2)

Obviously, Q will catch P when s(Q) = s(P)

Equating the above we obtain the quadratic equation

t^2 + (6-k)t + 2k = 0

For the above scenario to happen, the equation should have at least one real solution. And thus it's discriminant must be greater or equal to ZERO.

thus,

(6-k)^2 - 8k >= 0

or, k = 2 or 18.

While, k=2 is certainly a meaningless result, we accept k=18 as the answer!

The question asks us when $Q$ will catch up with $P$ . This means that the displacement $s$ should be constant at some point of time.

For the first dot $P$ , let us calculate the displacement.

Using the second law of motion,

$s = u_p t + \dfrac{1}{2} a_p t^{2}$ where $u_p$ denotes the initial velocity of the dot $P$ and $a_p$ denotes its acceleration.

Plugging in the known values we get, $s = 6t + t^{2}$

Now for the second object, the displacement is $s = u_q (t-2) + \dfrac{1}{2} a_q (t-2)^{2}$

$a_q = 0$ since the object $Q$ travels at constant speed. Thus, we get

$s = u_q (t - 2)$ for the dot $Q$

Since the displacement is equal, we can equate the two equations.

$u_q (t - 2) = 6t + t^2$ or

$u_q = \dfrac{6t + t^{2}}{t - 2}$

Thus, we have the initial velocity of the $Q$ object as a function of $t$ . We want to minimize $u_q$

Finding the minimum by taking the first two derivatives we get the minimum value of $u_q = 18$ at $t = 6$

Thus, the answer is $\boxed{18}$

Displacement of P in 2 sec from O would be:

using, s=ut + 0.5 a ${t^2}$ would be 16m

Now assuming P and Q meets at point Xm from O, and time taken to do so would be same.

For P:

X-16 = 10 t + ${t^2}$ ..............(1)

where initial velocity after 2 sec would be 10m/s using v = u+at

For Q:

k = X/t also X= kt.............(2)

Using equation 1 and 2

we form

(kt - 16) = 10t + ${t^2}$

k = 10 + t + 16 / t ........................(3)

Differentiating k wrt t we get

$\frac{dk}{dt}$ = 1 - 16 / ${t^2}$ = 0

we get t = 4

On double differentiation for max/min, we get min of k at t=4.

Hence substituting t=4 in equation 3

We get $\boxed{k=18m/s}$