Question

a body travelling with uniform accerlation crosses two points A and B with velocities 20 m/s and 80 m/s respectively if P and Q are the points on the line AB such that AP= PQ=QB then the ratio of velocity of the body at point p and Q is​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The ratio of velocity of the body at point p and Q is​ = √2 : √3

Explanation:

• Velocity at A point is initial velocity = u = 20 m/s.

       Velocity at B point is final velocity = v = 80 m/s.

    Suppose acceleration = a and distance between A, B = s

   From the 3rd law of motion, 2as = v² - u²

                                                    as = $\frac{v^2-u^2}{2}$ = $\frac{80^2-20^2}{2}$ = 600

• Velocity calculation at P point:

    Suppose at P point velocity = v'.

   As AP= PQ=QB and AB = s, so AP= PQ=QB = $\frac{s}{3}$

  For the motion from A to P point, initial velocity = u = 20 m/s.

  Final velocity = v', distance covered = $\frac{s}{3}$ and acceleration = a

  From 3rd law of motion, 2as = v² - u²

                                           $\frac{2as}{3}$ = v'² - 20²

                                             v'²= $\frac{2\times600}{3}$ + 400 [as as=600]

                                             v'² = 800

• Velocity calculation at Q point:

    Suppose at Q point velocity = u'.

  For the motion from Q to B point, initial velocity = u'.

  Final velocity = 80m/s, distance covered = $\frac{s}{3}$ and acceleration = a

  From 3rd law of motion, 2as = v² - u²

                                           $\frac{2as}{3}$ = 80² - u'²

                                             u'²=1600 -  $\frac{2\times600}{3}$  [as as=600]

                                             u'² = 1200

• Ratio calculation:  v'²:u'² = 800:1200 = 2:3

                                       v':u' = √2 : √3

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