Question 22: Motion in 1D

Question

The position of a particle related to time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$. The magnitude of velocity of the particle at $t=2 s$ will be :

Accepted Answer

The position of a particle as a function of time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$.
To find the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$, we first need to find the velocity of the particle as a function of time.
The velocity $v$ is the time derivative of the position $x$:
$$
v = \frac{dx}{dt}
$$
Taking the derivative of $x$ with respect to $t$, we get:
$$
v = \frac{dx}{dt} = 10t - 4\,\mathrm{m/s}
$$
Now we can find the velocity of the particle at $t=2\,\mathrm{s}$ by plugging in $t=2$:
$$
v(2\,\mathrm{s}) = 10(2) - 4\,\mathrm{m/s} = 16\,\mathrm{m/s}
$$
Therefore, the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ is:
$$
\boxed{|v(2\,\mathrm{s})| = 16\,\mathrm{m/s}}
$$

Answer

$14 \mathrm{~ms}^{-1}$

Answer

$16 \mathrm{~ms}^{-1}$

Answer

$10 \mathrm{~ms}^{-1}$

Answer

$06 \mathrm{~ms}^{-1}$

Motion in 1D

The position of a particle related to time is given by x=\left(5 t^{2}-4 t+5\right) \mathrm{m}. The magnitude of velocity of the particle at t=2 s will be :

Choose the correct answer: