Two motorists simultaneously drove out of cities A and B towards each other.

Two motorists simultaneously drove out of cities A and B towards each other. The first drove at a constant speed all the way. The second drove the first half at a speed of 70 km / h, and the second was 21 km / h faster than the speed of the first, as a result of which he arrived at B at the same time. first motorist. Find the speed of the first motorist.

Let us conditionally denote the entire path – s, the speed of the first motorist – x.

70 km / h – the speed of the second motorist when he was driving on the first part of the path.

х + 21 km / h – speed of the second motorist when he was driving the second part of the way.

t1 – the time of the first motorist spent on the road.

t2 – the time of the second motorist spent on the road.

t1 = t2 – the first motorist arrived at the same time as the second.

S = V * t.

t = S / V.

t – time, S – all the way, V – speed.

Then,

t1 – time of the first car, S – all the way, V – x.

t1 = S / x.

since the second motorist had one speed on one part of the road and another on the other, then S * ½ = S / 2.

t2 = (S / 2) / 70 + (S / 2) / (x + 21) = S / 140 + S / 2 * (x + 21).

Since t1 = t2.

S / x = S / 140 + S / 2 * (x + 21).

Reduce by S.

1 / x = 1/140 + 1/2 * (x + 21).

140 * (x + 21) / 140x * (x +21) = (x * (x +21) + 70x) / 140x * (x + 21).

140x + 21 * 140 = x2 + 21x + 70x.

140x + 2940 = x2 + 91x.

x2 + 91x – 140x – 2940 = 0.

x2 – 49x – 2940 = 0.

D = (- 49) 2 – 4 * 1 * (- 2940) = 2401 + 11760 = 14161.

The root of 14161 = 119.

x1 = (49 – 119) / 2 * 1 = – 70/2 = – 35.

x2 = (49 + 119) / 2 * 1 = 168/2 = 84.

The first root does not satisfy the condition, so the speed of the first motorist is 84 km / h.

Solution: 84 km / h.