Question on time speed and distance A and B, who are separated by a distance 90 m, are approaching towards each other. The initial speed of A is 5 m/s and that of B is 10 m/s. If A increases his speed by 5 m/s every second, when will they meet each other?
HINT:

Let the distance $d$ between the two be expressed a a function of time $t$. Then you have that the rate of change of the distance is 
$$d'(t)=-5-10-5t$$
$$d'(t)=-15-5t$$
The two will meet when $d(t)=0$. Can you integrate, find the constant that is added when you integrate, and then find when $d(t)=0$?

FULL ANSWER:
Integrate both sides to get
$$d(t)=-15t-\frac{5}{2}t^2+C$$
And since the starting distance is $90$, then $C=90$, so
$$d(t)=-15t-\frac{5}{2}t^2+90$$
And you must find when this is zero:
$$-\frac{5}{2}t^2-15t+90=0$$
$$\frac{5}{2}t^2+15t-90=0$$
This can be solved like a quadratic:
$$t^2+6t-36=0$$
Can you use the quadratic formula to solve this?
$A$ &amp; $B$ are $90$ m distance apart.

Their relative speed is $(10+5)$ m/sec

Say, after $t$ second, they meet each other at a distance $x$ from the point from where $A$ starts.

We can say that both of them in combination have covered $90$ m distance within $t$ secs at a speed of $15$ m/s

so $t = \frac{90}{15} = 6$ secs