The speed of a particle moving in a circle of radius 0.1m is v = 1.0t where t is time in second. The resultant acceleration of the particle at t = 5s will be

The correct option is 250 m / s².

To find the resultant acceleration of a particle moving in a circular path, we need to consider both the tangential acceleration ($a_t$) and the radial (centripetal) acceleration ($a_c$).

Step 1: Calculate the tangential acceleration ($a_t$)
The speed of the particle is given as a function of time:
$v=1.0t$
Tangential acceleration is the rate of change of speed with respect to time:
$a_t=\frac{dv}{dt}=\frac{d}{dt}\left(1.0t\right)=1.0{\text{m/s}}^2$

Step 2: Calculate the radial (centripetal) acceleration ($a_c$) at $t=5\text{s}$
First, find the speed of the particle at $t=5\text{s}$:
$v=1.0\times 5=5\text{m/s}$
The radius of the circular path is given as:
$r=0.1\text{m}$
Now, calculate the radial acceleration using the formula:
$a_c=\frac{v^2}{r}$
Substituting the values at $t=5\text{s}$:
$a_c=\frac{5^2}{0.1}=\frac{25}{0.1}=250{\text{m/s}}^2$

Step 3: Calculate the resultant acceleration ($a$)
Since the tangential acceleration and radial acceleration are perpendicular to each other, the magnitude of the resultant acceleration is given by:
$a=\sqrt{a_t^2+a_c^2}$
Substitute the values of $a_t$ and $a_c$:
$a=\sqrt{{1.0}^2+{250}^2}=\sqrt{1+62500}=\sqrt{62501}\approx 250{\text{m/s}}^2$

Since $a_t$ is extremely small compared to $a_c$, the contribution of $a_t$ to the resultant acceleration is negligible. Thus, the resultant acceleration is approximately equal to the radial acceleration:
$a\approx 250{\text{m/s}}^2$