Question

an object accelerates from rest to a velocity 27.5 metre per second in 10 second then find the distance covered by object in next 10 second​

Answer 1

Answer:

Explanation:

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Answer 2

An object accelerates from rest to a velocity 27.5 m/s in 10 second.

Given that, the initial velocity of the object is 0 m/s (as it starts from rest), final velocity is 27.5 m/s and time is 10 seconds.

We have to find the distance covered by object in next 10 second.

Using the First Equation Of Motion,

v = u + at

Substitute the known in above formula,

→ 27.5 = 0 + a(10)

→ 27.5 = 10a

→ 27.5/10 = a

→ 2.75 = a

Therefore, the acceleration of the object is 2.75 m/s².

Now,

Using the Second Equation Of Motion,

s = ut + 1/2 at²

In next 10 seconds, initial velocity becomes 27.5 m/s.

Substitute the known in the above formula,

→ s = 27.5(10) + 1/2 × 2.75 × (10)²

→ s = 275 + 1/2 × 2.75 × 100

→ s = 275 + 1/2 × 275

→ s = 275 + 137.5

→ s = 412.5

Therefore, the distance covered by the object in next 10 second is 412.5 m.