The speed of a car is v km / h, the distance traveled by t h is (algebraic) km?

The speed of a car is v km / h, the distance traveled by t h is (algebraic) km?

The speed of a car is v km / h, the distance traveled by t h is (VT) km

The distance between a and B is 100 km, and the driving speed of a car is v km / h. (1) the algebraic expression is used to express the driving time of the car from a to B; (2) if the speed increases by 5 km / h, how long does it take? How long will it be earlier than before? (3) when v = 50 km / h, calculate the above algebraic expressions

According to the meaning of the title, we get (1) 100V = 100 / V (hours), when v = 50 km, 100V = 10050 = 2 (hours); (2) 100V + 5 hours, when v = 50 km, 100V + 5 = 10050 + 5 = 2011 (hours); ② (100V − 100V + 5) hours, when v = 50 km, 100V − 100V + 5 = 10050 − 10055 = 211

When a freighter runs between terminals a and B, it is known that the sailing speed of the ship in still water remains unchanged, which is a km / h, and the water flow speed is 3 km / h. The average speed of the ship in one round trip is expressed by algebraic formula
Don't copy it. Why 1 / A-3 + 1 / A + 3

This is simple. The ship is affected by the current in the water. A is the speed of the ship relative to the water, so the speed (absolute speed) is a + 3 and A-3, and the average speed is the ratio of the total distance and time, the time of two times s / (a + 3) and S / (A-3), so t = T1 + T2, v = 2S / T, so v = 1 / (1 / (a + 3) + 1 / (A-3)), your answer is not right

The speed of a ship in still water is a km / h, the speed of water flow is B km / h, a > B. the distance between a and B docks is C km / h

TDO = s △ V1 = CKM ^ (AKM / H + BKM / h)
T loop = s △ V2 = CKM △ AKM / h-bkm / h
T round trip = T1 + T2 = (C / A + b) H + (C / a-b) H = 2Ac / a square + b square