The cart traveled for the first third of the way at a speed of V1 = 5 km / h, for the rest of the way it traveled at a speed

The cart traveled for the first third of the way at a speed of V1 = 5 km / h, for the rest of the way it traveled at a speed twice as high as the average speed along the entire route. Find the speed of the cart on the second part of the path.

Decision:
1. Average speed – the ratio of the total distance traveled to the entire time of movement.
2. Let’s designate: 3s – all the way covered by the cart; U2 – speed on the second part of the track.
3. The travel time on the first part of the path is s / 5. The travel time on the second part of the path is 2s / U2. The total travel time of the cart is:
s / 5 + 2s / U2 = (s * U2 + 10s) / (5 * U2);
4. Average speed is equal to:
3s: ((s * U2 + 10s) / (5 * U2)) = (3s * 5 * U2) / (s * U2 + 10s) = 15U2 / (U2 + 10);
By condition, the average speed is half the speed on the second part of the path. The equation turns out:
U2 / 2 = 15U2 / (U2 + 10);
30U2 / (U2 + 10) = U2;
30 / (U2 + 10) = 1;
U2 + 10 = 30;
U2 = 30 – 10;
U2 = 20;
Answer: the speed on the second part of the path is 20 km / h.