Aparticleis movingin acircleofradiusR withconstantspeed. Thetimeperiodof particle is T=1 second. In atime t =T/6,if the difference betweenavg. speedandmagnitude ofavg. velocityis 2m/s,find radiusofthecircleinmetres.

.:Here the Body is moving in a roundabout way of sweep 21/22 => the circuit of the circle is 2*pi*r = 2*(22/7)*(21/22)=6. 

in this manner the Body takes 6 seconds to venture to every part of the entire roundabout way. Presently in 2 seconds, it will go to a Point on the circle which makes an edge 120 degrees with the Initial Position about the middle. 

In this manner the uprooting of the Body is = the separation amongst Initial and last Points. It can be ascertained (utilizing some trignometry) as r*sqrt(3) = 21/22 * sqrt(3). 

also, the time is 2 sec => the normal speed = sqrt(3) * (21/22)/2 

Also, the normal quickening is characterized as the vectorial distinction amongst last and Initial speeds after some time. The vectorial distinction for this situation gives the Magnitude as 1 and the time is 2 sec, which means the normal speeding up = 0.5 

A normal estimation of an amount is characterized as the distinction of Initial and last estimations of it's chance differential after some time. If there should be an occurrence of speed it is the distinction of Position (time differential of speed) after some time. Also, for speeding up, it is contrast in speed after some time. What's more, for this situation the amounts are vectors, so it would be vectorial distinction.