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

Answer:

Given,

Acceleration, a=36km/h

2

Initial velocity, u=12km/h

Final velocity, v=18km/h

Using the equation of motion as:

v

2

−u

2

=2as

Where, s is distance traversed by ship.

(18)

2

−(12)

2

=2(36)s

324−144=36s

s=5km

Distance traversed by ship is 5 Km

Answer 2

Answer:

The distance traveled by the ship during this period is 2.5 km.

Explanation:

As per third equation of motion,

\bold{v^{2}=u^{2}+2 a s arrow(1)}v

2

=u

2

+2asarrow(1)

Here acceleration, a = 36 Km/hour^{2}hour

2

Initial velocity, u = 12 km/hour

Final velocity, v = 18 km/hour

S = distance traveled by the ship in Km

Hence, 2 a s=\bold{v^{2}-u^{2}}v

2

−u

2

2as = (18)^{2}-(12)^{2}(18)

2

−(12)

2

= 324 – 144 = 180

s = 180 / 2a = 180 / 2 x 36 = 2.5 Km

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