Referring to $v - s$ diagram, find:
(a) acceleration of the particle when its velocity becomes half of the initial velocity.
(b) total distance covered by the particle
(c) average velocity of the particle during its total time

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Solution

(i)
$a = v (\frac{d v}{d s})$
$= 20 \times (\frac{20 - 0}{0 - 50})$
$= 20 \times \frac{20}{50}$
$= - 8 m / s^{2}$

(ii)
$v^{2} = u^{2} - 2 a s$
$0 = u^{2} - 2 a s \Rightarrow \frac{400}{2 \times 8} = s$
$25 = s$

(iii)
$t = \frac{20 - 0}{8} = \frac{20}{8}$
$V a l o c i t y = \frac{change in distance}{time}$
$= \frac{25}{20} \times 8 = 10 m / s$

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Referring to v−s diagram, find:(a) acceleration of the particle when its velocity becomes half of the initial velocity.(b) total distance covered by the particle(c) average velocity of the particle during its total time

(i)a=v(dvds)=20×(20−00−50)=20×2050=−8m/s2(ii)v2=u2−2as0=u2−2as⇒4002×8=s25=s(iii)t=20−08=208Valocity=changeindistancetime=2520×8=10m/s

Referring to $v - s$ diagram, find:
(a) acceleration of the particle when its velocity becomes half of the initial velocity.
(b) total distance covered by the particle
(c) average velocity of the particle during its total time