Question

An object moves on a circular path such that the distance covered is given by S = (0.5t + 2t) m, here t is time in s. If
radius of circle is 4 m, then initial acceleration of object is
5​

Answer 1

Given:

A object moves on a circular path such that the distance covered is given by S=(0.5t^2 + 2t)m here t is time in seconds. The radius of circle is 4m.

To find:

Initial acceleration of the object ?

Calculation:

Let tangential acceleration be a_(t):

$\therefore \: s = 0.5 {t}^{2} + 2t$

$\implies \: v = \dfrac{ds}{dt}$

$\implies \: v = \dfrac{d(0.5 {t}^{2} + 2t) }{dt}$

$\implies \: v = t + 2$

$\implies \: a_{t} = \dfrac{dv}{dt}$

$\implies \: a_{t} = \dfrac{d(t + 2)}{dt}$

$\implies \: a_{t} = 1 \: m {s}^{ - 2}$

So, tangential acceleration initially is 1 m/s².

Let centripetal acceleration be a_(c):

$\therefore \: a_{c} = \dfrac{ {v}^{2} }{r}$

$\implies \: a_{c} = \dfrac{ {(t + 2)}^{2} }{r}$

Putting t = 0 secs for initial value:

$\implies \: a_{c} = \dfrac{ {(0 + 2)}^{2} }{4}$

$\implies \: a_{c} = \dfrac{ 4}{4}$

$\implies \: a_{c} = 1 \: m {s}^{ - 2}$

So, net acceleration:

$\therefore \: a_{net} = \sqrt{ {(a_{t})}^{2} + {(a_{c})}^{2} }$

$\implies \: a_{net} = \sqrt{ {1}^{2} + {1}^{2} }$

$\implies \: a_{net} = \sqrt{2 } \: m {s}^{ - 2}$

So, net acceleration is √2 m/s².