Question

It is known that from t = 2 s to t = 10 s the acceleration of a paurticle is inversely the time t. When t = 2 s, v = -15 m/s, and when t = 10s, proportional to the cube v = 0.36 m/s. Knowing that the particle is twice as far from the origin when t = 2s as it is when t = 10 s, determine (a) the position of the particle when t = 2 s and when t = 10 s, (b) the total distance traveled by the particle from t = 2 s to
t = 10 s.​

Answer 1

Explanation:

v = u+at

0.36 = -15+a(10-2)

15.36 = 8a

a= 15.36/8

= 1.92 m/s²

s = ut + at²/2

= 0.36×8+1.92×64

= 28.8+122.88

= 151.68 meter

Hope this helps you

Answer 2

Answer:

The position of particle= 1.92 m/s²
The total distance of particle = 151.68 meter

Explanation:
           u= initial velocity of the body

           v= final velocity of the body

           a= acceleration

           t= time taken
           S=distance

A) Position of particle

          $v = u+at$

         0.36  = -15+a (10-2)

         15.36 = 8a

                   = 15.36/8

                a  = 1.92 m/s²

B) Distance of particle

            $s = ut + at^{2} /2$

            = 0.36×8+1.92×64

            = 28.8+122.88

         s = 151.68 meter

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