Predict a Collision: Will the robber get away? Think of a road as a number line, starting at 0 and continuing indefinitely:
.................................................................

There are two cars on the road: C and R. C is the cop who is trying to catch R, the robber. C starts at 0, and R starts somewhere on the road:
C.............................R..................................

The cop is already moving - he's chasing the robber. He has a constant speed. The robber just hopped into his car. He's accelerating. Each tick, the robber's speed increases by his acceleration.
Say the cop's speed is 7 and the robber's acceleration is 1. If the robber starts at 30, this is what the road would look like each tick:
C.............................R..................................
.......C.......................R.................................
..............C..................R...............................
.....................C..............R............................
............................C...........R........................
...................................C.........R...................
..........................................C........R.............
.................................................C........R......

After the last tick above, the robber's speed is equal to the cop's, and he's still ahead. Since the cop is moving at a constant speed and the robber is still speeding up, the robber escapes, so you output a truthy value. However, if the cop's speed had been 9...
C.............................R..................................
.........C.....................R.................................
..................C..............R...............................
...........................C........R............................
....................................C...R........................
.............................................X...................

... then the cop catches up to the robber before the robber can get away (marked by the X), so you output a falsey value.
Your Task
Given three inputs - the cop's speed, the robber's position, and the robber's acceleration - determine whether or not the robber will get away.
Rules

The cop always starts at 0.
All inputs will be positive integers.
The cop catches the robber if, after any tick, the cop's position is greater than or equal to the robber's position.
The robber gets away when he hasn't been caught yet and his speed is greater than the cop's.
Your program must terminate after output.
The robber accelerates before he moves each tick.

Test Cases
Cop Speed, Robber Position, Robber Acceleration -&gt; Output


7, 30, 1 -&gt; truthy
9, 30, 1 -&gt; falsey
2, 1, 3 -&gt; truthy
100, 100, 50 -&gt; truthy
60, 60, 20 -&gt; falsey
10, 1, 1 -&gt; falsey
10, 50, 2 -&gt; truthy
11, 50, 2 -&gt; truthy
12, 50, 2 -&gt; truthy
13, 50, 2 -&gt; truthy
14, 50, 2 -&gt; truthy
15, 50, 2 -&gt; truthy
16, 50, 2 -&gt; falsey
17, 50, 2 -&gt; falsey
18, 50, 2 -&gt; falsey
100, 451, 10 -&gt; truthy

Reference Python 3 implementation that creates a visual also: Try it online!
This is code-golf, so shortest answer in bytes wins.
Python 3, 29 bytes



lambda s,p,a:(a-2*s)**2&lt;8*a*p


Try it online!

Explanation

The cop's position at time t is st.

The robber's position at time t is a(t)(t+1)/2 + p.

The signed distance from the cop to the robber is (a/2)t^2 + (a/2-s)t + p.

It never reaches zero if the discriminant is negative, the discriminant being (a/2 - s)^2 - 4(a/2)(p) = [(a-2s)^2-8ap]/4, which has the same sign as (a-2s)^2-8ap.