Question

A point moves along a circle with speed,p=dm,{d=0.5 m/s^2}...p,d,m are just variables!!
Find the total speed and acceleration of point at moment when it covers (1/10)th of the circle after beginning of motion..!!!(Non-Uniform Motion)

Answer 1

@april115 @andriod09 @AnElephant @ayushkr2611 @blueocean18 @countonme123 @Coolsector @dan815 @e.cociuba @Falco276 @Fifciol @GoldEverything @GOODMAN @HaperFink22 @ilfy214 @ivancsc1996 @Luigi0210 @Loser66 @lncognlto

Answer 2

If it is a uniform circular motion then d. We assume p is a speed and m is time

Answer 3

okay

Answer 4

Then???

Answer 5

The answer is d since the acceleration is constant. Maybe you are thinking about the total speed?

Answer 6

O_O....Oopsy!!

Answer 7

Yes speed

Answer 8

Then, we can use this equation: \[\theta _{f}=\theta _{i}+\omega _{i} t+\frac{ 1 }{ 2 } \alpha t ^{2}\]Where theta is the angle, omega is the angular velocity and alpha is the angular acceleration. Since initial angle and the initial angular velocity are zero, when we clear t we get:\[\sqrt{\frac{ 2 \theta _{f} }{ \alpha }}=t\] Now, alpha is related to d (the linear acceleration) by:\[\alpha=\frac{ d }{ r }\]That means that you need the radius of the circle. Do you have it?

Answer 9

r??

Answer 10

t,then will equal \[2\sqrt{{\pi*r}/5}\]

Answer 11

@ivancsc1996

Answer 12

so v wil be \[\sqrt{{(pi*r)}/5}\]

Answer 13

No, \[m=\sqrt{\frac{ 2 \times \frac{ 2 \pi }{ 10 }}{ \frac{ d }{ r } }}=\sqrt{\frac{ 2 \pi r}{ 5d }}\]Then you substitute that back into the equation for p and you get:\[p=d \sqrt{\frac{ 2 \pi r }{ 5d }}=\sqrt{\frac{ 2 d \pi r }{ 5 }}\]

Answer 14

It shud be 4*2x/10

Answer 15

@april115 @andriod09 @AnElephant @ayushkr2611 @blueocean18 @countonme123 @Coolsector @dan815 @e.cociuba @Falco276 @Fifciol @GoldEverything @GOODMAN @HaperFink22 @ilfy214 @ivancsc1996 @Luigi0210 @Loser66 @lncognlto

Answer 16

@androidonyourface @ash2326 @amistre64 @surjithayer @sorour

Answer 17

@TuringTest