
A boat takes 2 hours to go 8 km and come back to a still water lake. With velocity of 4 km/hr the time taken for going upstream 8 km and going back is
A 140 minutes
B 150 minutes
C 160 minutes
D 170 minutes
Answer
232.8k+ views
Hint: As the boat goes to 8 km and comes back so we have to add both distances one for going and another is for coming and then use the proper formula for speed, distance and time.
Complete step-by-step answer: It is given that the boat goes 8 km and comes back in still water and it takes 2 hours for this. So we first find the velocity of boat Vb
As velocity is distance covered by the boat per unit time,
Distance covered by boat in 2 hours is = 16
Therefore, velocity of boat is
⇒ ${V_b} = \dfrac{{8 + 8}}{2} = \dfrac{{16}}{2} = 8km/hr$
And velocity of water is Vw = 4km/hr
Now, time taking in going upstream is distance covered by boat in going 8 km divided by the velocity of upstream and upstream velocity is the difference of the velocities of boat and water
⇒ Time taken for going upstream is ${t_1} = \dfrac{8}{{{V_b} - {V_w}}} = \dfrac{8}{{8 - 4}} = 2hr = 2 \times 60\min = 120\min $
Now, Time taken by boat in going downstream is distance covered by boat in going 8 km back divided by the speed of downstream and speed of downstream is the sum of the velocity of the boat and velocity of water
⇒ Time taken for going downstream is ${t_2} = \dfrac{8}{{{V_b} + {V_w}}} = \dfrac{8}{{8 + 4}} = \dfrac{8}{{12}}hr = \dfrac{8}{{12}} \times 60\min = 140\min $
As the water helps the motion of boat then total time taken by boat is sum of time taken by boat to go upstream and the time taken by boat to go downstream
⇒ Total time taken by boat is $t = {t_1} + {t_2} = 120 + 40 = 160\min $
Hence, C option is correct.
Note: Here keep in mind that when the boat is going downstream the boat opposes the velocity of water so we add there velocities and while going upstream then water and boat are in the same direction so we take the difference between there velocities.
Complete step-by-step answer: It is given that the boat goes 8 km and comes back in still water and it takes 2 hours for this. So we first find the velocity of boat Vb
As velocity is distance covered by the boat per unit time,
Distance covered by boat in 2 hours is = 16
Therefore, velocity of boat is
⇒ ${V_b} = \dfrac{{8 + 8}}{2} = \dfrac{{16}}{2} = 8km/hr$
And velocity of water is Vw = 4km/hr
Now, time taking in going upstream is distance covered by boat in going 8 km divided by the velocity of upstream and upstream velocity is the difference of the velocities of boat and water
⇒ Time taken for going upstream is ${t_1} = \dfrac{8}{{{V_b} - {V_w}}} = \dfrac{8}{{8 - 4}} = 2hr = 2 \times 60\min = 120\min $
Now, Time taken by boat in going downstream is distance covered by boat in going 8 km back divided by the speed of downstream and speed of downstream is the sum of the velocity of the boat and velocity of water
⇒ Time taken for going downstream is ${t_2} = \dfrac{8}{{{V_b} + {V_w}}} = \dfrac{8}{{8 + 4}} = \dfrac{8}{{12}}hr = \dfrac{8}{{12}} \times 60\min = 140\min $
As the water helps the motion of boat then total time taken by boat is sum of time taken by boat to go upstream and the time taken by boat to go downstream
⇒ Total time taken by boat is $t = {t_1} + {t_2} = 120 + 40 = 160\min $
Hence, C option is correct.
Note: Here keep in mind that when the boat is going downstream the boat opposes the velocity of water so we add there velocities and while going upstream then water and boat are in the same direction so we take the difference between there velocities.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

