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EXPLANATION.

Cost price of book = 16.

Equal to selling price of 20 books.

As we know that,

Loss % = [loss/cp] x 100.

Using this formula in this question, we get

Let, cp of one book = x.

Cp of 20 books = 20x.

Loss = 20x - 16x = 4x.

Loss % = (4x)/(20x) x 100.

Loss % = 4 x 5 = 20%.

Loss % = 20 %.

(2) Acceleration of a particle moving in a straight line at any instant point t.

a = 2t + 3 m/s.

To find velocity we integrate the expression, we get.

[tex]\sf \displaystyle dv = a.dt[/tex]

[tex]\sf \displaystyle \int\limits^v_0 dv = \int\limits^2_0 a .dt[/tex]

[tex]\sf \displaystyle \int\limits^v_0 dv = \int\limits^2_0 (2t + 3) dt[/tex]

[tex]\sf \displaystyle \int\limits^v_0 dv = \int\limits^2_0 2t .dt + \int\limits^2_0 3.dt[/tex]

[tex]\sf \displaystyle v = \bigg[\frac{2t^{2} }{2} \bigg]_{0}^{2} + \bigg[ 3t \bigg]_{0}^{2}[/tex]

[tex]\sf \displaystyle v = \bigg[t^{2} \bigg]_{0}^{2} + \bigg[ 3t \bigg]_{0}^{2}[/tex]

[tex]\sf \displaystyle v = [(2)^{2} - (0)^{2} ] + [ 3(2) - 3(0)][/tex]

[tex]\sf \displaystyle v = 4 + 6[/tex]

[tex]\sf \displaystyle v = 10 \ m/s[/tex]

Velocity of particles at t = 2 seconds = 10 m/s.

To find displacement.

Integrate the expression of velocity.

v = t² + 3t.

[tex]\sf \displaystyle x = \int\limits^2_0 (t^{2} + 3t)dt[/tex]

[tex]\sf \displaystyle x = \int\limits^2_0 t^{2} dt + \int\limits^2_0 3t dt[/tex]

[tex]\sf \displaystyle x = \bigg[ \frac{t^{3} }{3} \bigg]_{0}^{2} + \bigg[ \frac{3t^{2} }{2} \bigg]_{0}^{2}[/tex]

[tex]\sf \displaystyle x = \bigg[\frac{(2)^{2} }{3} - 0 \bigg] + \bigg[ \frac{3(2)^{2} }{2} - 0 \bigg][/tex]

[tex]\sf \displaystyle x = \frac{8}{3} + 6[/tex]

[tex]\sf \displaystyle x = \frac{8 + 18}{3} = \frac{26}{3} \ m[/tex]

Displacement of particles at t = 2 seconds = 26/3 m.