Question

The speed of a particle moving in a circle of radius r=2m varies witht time t as v=t^(2), where t is in second and v in m//s. Find the radial, tangential and net acceleration at t=2s.

Answer 1

The radial, tangential and net acceleration at t=2s are 8 m/s², 4 m/s² and √80 m/s².

- Linear speed of particle at t = 2 s is

v = (2)² = 4 m/s

- Radial acceleration  is given  by :

$a_{r}$ =  v²/r

- Radial acceleration  at t=2s is

$a_{r}$ = (4)²/2 = 16/2 = 8 m/s²  

- Tangential acceleration is given by :

$a_{t}$ =  dv/dt = 2t

∴ Tangential acceleration at t=2s is

$a_{t}$ =2 ×2 = 4 m/s²

∴ Net acceleration of particle t=2s is

a =  √($a_{r}$)² + ($a_{t}$)²

a = √(8)² + (4)²  

a = √64 + 16

a = √80 m/s²

Answer 2

Given:

Radius of circle $(r)= 2m$

Speed of a particle $(v) = t^2 m/s$

To find: The radial, tangential and net acceleration at $t=2s$.

Solution:

Linear speed of particle at     $t=2s$  is

            $v = (2)^2 =4m/s$

Radial acceleration  $(a_r)= \frac{v^2}{r}$

                                 $a_r = \frac{(4)^2}{2} \\a_r = 8 m/s^2$

So, Radial acceleration is $= 8m/s^2$

The tangential acceleration is $a_t= \frac{dv}{dt}$

​​                                                $a _t = 2t$

∴ Tangential acceleration $t=2s$ is

                                     $a _t =(2)(2)\\a _t=4m/s$

 ∴ Net acceleration of particle $t=2s$ is

                             $a = \sqrt{a_r^2+ a_t^2} \\ \\a =\sqrt{8^2+4^2}$

                             $a= \sqrt{80}m/s^2$

Thus net acceleration of particle is   $a= \sqrt{80}m/s^2$