Question

Ship moving with constant acceleration of 36 km per hour square in a fixed direction speeds up from 18 km per hour to 36 km per hour. Find the distance travelled by the ship in this period. ​

Answer 1

Given:-

• Initial velocity (u) = 18km/h

• Final velocity (v) = 36km/h

• Acceleration (a) = 36km/h²

To Find:-

• Distance covered by the ship (s)

Solution:-

Firstly we convert the unit here

Initial velocity = 18×5/18 = 5m/s

Final Velocity = 36km/h = 36×5/18 = 10m/s

Acceleration = 36×5/18 = 10m/s²

Now using 3rd equation of motion

→ v² = u² +2as

Substitute the value we get

→ 10² = 5² + 2×10×s

→ 100 = 25 + 20×s

→ 100 -25 = 20×s

→ 75 = 20×s

→ s = 75/20

→ s = 3.75 m

∴ The distance covered by the ship is 3.75m.

Answer 2

Given :-

• Acceleration (a) ↪36km/h² ↪36× 5/18 ↪10m/s².
• Initial velocity (u) ↪18km/h ↪18 × 5/18 ↪5m/s.
• Final velocity (v) ↪36km/h ↪36 × 5/18 ↪10m/5s.

To Find :-

• The distance (s) travelled by ship.

Solution :

☀ The distance travelled by ship. ☀

↪ Applying 3rd law of motion, We get :

↪ v² = u² + 2as

↪ 10² = 5² + 2 × 10 × s

↪ 100 = 25 + 20 + s

↪ 100 = 45 + s

↪s = 100 - 45

↪s = 55m.

.°. Distance is 55m.