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
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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.