Question

A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is `5ms^(-1)` and the speed is increasing at a rate of `2ms^(-2)`. At this instant, the magnitude of the net acceleration will be
A. `3.2ms^(-2)`
B. `2ms^(-2)`
C. `1.2ms^(-2)`
D. `4.3ms^(-2)`

Answer 1

Correct Answer - A
`a=sqrt(a_(t)^(2)+a_(n)^(2))`
`a_(t)` = rate of change of speed =`2 ms^(-2)`
`a_(n)=(v^(2))/(R)=((5)^(2))/(10)=2.5ms^(-2)`
`therefore" "a=sqrt(a_(t)^(2)+a_(n)^(2))=sqrt((2)^(2)+(2.5)^(2))=3.2ms^(-2)`