Question

a ship moves with a constant acceleration of 36 km per hour squared.Its velocity changes from 12km per hour to 18 km per hour .Find the distance travelled by the ship during this period.

Answer 1

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Answer 2

Answer:

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

Explanation:

As per third equation of motion,  

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

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

Initial velocity, u = 12 km/hour

Final velocity, v = 18 km/hour

S = distance traveled by the ship in Km

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

2as = $(18)^{2}-(12)^{2}$ =  324 – 144 = 180

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