Rapid physics

In the first case speed of water=1 m/s,so volume of water flows through 60m wide and 2m deep cross-section in 1 second=1x60x2 m3=120 m3...(1) Let,speed of water is t m/s where cross-section of the river is 10m wide and 2m deep. Here volume of water flows in 1 second is tx10x2 m3=20t m3...(2) Now as (1) & (2) is equal, hence we get, 20t m3=120 m3 or,t=6 hence,speed of water where river narrows to be 10m wide(but is still 2m deep) is 6 m/s.

By the equation of continuity, a1v1=a2v2. Hence 60x2x1=10x2xv2. Therefore v2=6m/s.

Easy.Just multiply by 6

According to continuity equation, the volume flow rate, Q is constant. So, Q=AV=av, where A and V represent cross sectional area and speed of river at one point, a and v represent cross sectional area and speed of river at another point.

AV=av

(60x2)1=(10x2)v

v=6 m/s

Assuming like conditions at both points in the river, we know that the volume of water passing per second at the first point must be equal to the volume of water passing per second at the second point.

At the first point $60*2*1$ $m^3$ passes per second. At the second point $10*2*answer$ $m^3$ passes per second.

Equating and solving gives answer = 6

From the equation of continuity , we know that , A1v1=A2v2 , let's assume A1 as the previous area of the river and v1 as the previous velocity of the water and just alike let's assume A2 as the new area of the river and v2 as the new velocity(after the shrink of the width of the river) . So , from the equation , we can get that , v2= A1v1/A2 , =>6 m/s .