Question

hi there!!

$\sf\huge{\underline{\mathbb{QUESTION}}}$

$<b><h>A cyclist starts from the centre of a circular park of radius 1 Km , reaches the edge 'P' of the park , then cycles along the circumference , and returns to the centre along QO as shown in the fig. If the round trip takes 10 minutes, what is the (a) Net displacement (b) average velocity (c) Average speed$​

Answer 1

$\sf\huge{\underline{\mathbb{Answer}}}$

$\sf\bold{\underline{\mathfrak{Question!!}}}$

$<b><h>A cyclist starts from the centre of a circular park of radius 1 Km , reaches the edge 'P' of the park , then cycles along the circumference , and returns to the centre along QO as shown in the fig. If the round trip takes 10 minutes, what is the (a) Net displacement (b) average velocity (c) Average speed$.

$1.Total displacement$=0, since he returns to the same place or Initial position , so the net displacement will be 0.

$2.Average Velocity:$ since total displacement is zero, hence the velocity will be zero.

$3.Firstly , he travels a distance R i.e, OP, then move along the circumference of circle Of distance QP,where he again move towards the centre of circle through OQ of distance R</p><p></p><p>hence total distance covered is..</p><p></p><p>R + R + [tex]\frac{2R}{4}$π =S

S = 2R+$\frac{R}{2}$π

S= 2R + 1.57R

S = 3.57R...

S =$3.57\times 1$

S = 3.57km..

time, t= 10min= $\frac{1}{6}$

t = $\frac{1}{6}$

SPEED = $\frac{DISTANCE}{TIME}$

Speed = $\frac{3.57}{1/6}$

= $3.57\times 6$

=> 21.42km\h (or) 5.95m/s

Therfore , speed of the cyclist is 5.95m/s (or) 21.42km\h..