Question

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The velocity of a body changes from 65 m/s - 98 m/s by Time of 12 s . What is distance covered by body if acceleration is 2.75 m/s2.​​

Answer 1

According to the Question

It is given that

• Initial velocity ,u = 65m/s

• Final velocity ,v = 98m/s

• Time taken ,t = 12s

• Acceleration ,a = 2.75m/s²

We have to calculate the distance covered by body.

we will use here kinematics equation of motion.

• v² = u² + 2as

Where v is the final velocity

a is the acceleration

u is the initial velocity

v is the final velocity

s is the displacement of the body

Substitute the value we get

=> 98² = 65²+ 2*2.75*s

=> 98² - 65² = 2*2.75*s

=> (98-65)(98+65) = 5.5 *s

=> 33*163 = 5.5 × s

=> 5379 = 5.5 ×s

=> s = 5379/5.5

=> s = 978m

Hence, the distance covered by the body is 978 metres.