Question

The velocity of a boat in still water is 10 m/s. If water flows in the river with a velocity of 6m/s what is the difference in times taken to cross
the river in the shortest path and the shortest time. The width of the river is 80 m.
1) 1 sec
2) 10 sec
3)√3/2 sec 4) 2 sec​

Answer 1

Answer:

15 second

Explanation:

velocity of the boat in still water is 10m/s

Velocity of river is 6m/s

so, velocity of river while coming downwards is 10+6=16 m/s

And velocity of the boat while going upwards is 10-6=4m/s

displacement is 80m

for first case

distance=speed ×time

80=16t

t =5 second

for second case

d=s×t

80=4t

t=20sec

time difference=20-5

15 second

which does not match any option so please make sure that all the details of your question is correct and verify it.

Answer 2

Given: the velocity of the boat in still water, v₁ = 10 m/s

           the velocity of the river water, v₂ = 6 m/s

           the width of the river, d = 80 m

To Find: the difference in time taken, Δt = t₁ - t₂.

Solution:

To calculate Δt, the formula used:

• time = distance / velocity
• time = width of the river / velocity
• t = d / V

Applying the above formula:

For the shortest path:

t₁ = d / V₁

V₁ = √v₁ - v₂

∴ t₁ = d / √v₁² - v₂²

     = 80 / √10² - 6²

     = 80/ √100 - 36

     =  80/ √64

     =  80 / 8

     = 10

t₁    = 10 s                                                              ⇒ 1

For the shortest time:

t₂ = d / V₂

here, V₂ = speed of boat in still water i.e. v₁

∴ t₂ = d / v₁

      = 80 / 10

      = 8 / 1

      = 8

  t₂ = 8 s                                                              ⇒ 2

On subtracting equations 1 and 2:

Δ t  = t₁ - t₂

     =  10 - 8

     =  2

Δt  = 2 s

Hence, the difference in the time taken is 2 seconds (option-4).