The position of a particle is given by: r =3.0t i -2.0t^2 j +4 k … - Vidyadip

Question

The position of a particle is given by: $\vec{\text{r}}=3.0\text{t}\hat{\text{i}}-2.0\text{t}^2\hat{\text{j}}+4\hat{\text{k}}\text{ m}$ where $t$ is in seconds, $r$ is in metres and the coefficients have the proper units.

Find the velocity $v$ and acceleration $a.$
What is the magnitude of velocity of the particle at $t = 2s$?

✓

Answer

$\vec{\text{r}}=3.0\text{t}\hat{\text{i}}-2.0\text{t}^2\hat{\text{j}}+4\hat{\text{k}}\text{ m}$

$\vec{\text{v}}=$ velocity $=\frac{\text{d}\vec{\text{r}}}{\text{dt}}=3.0\hat{\text{i}}-4.0\text{t }\hat{\text{j }}\text{m/s}$

$\vec{\text{a}}=$ acceleration $=\frac{\text{d}\vec{\text{v}}}{\text{dt}}=-4.0\hat{\text{ j }}\text{m/s}^2$

Magnitude of velocity at $t = 2s$

$\vec{\text{v}}=3.0\hat{\text{i}}-8.0\hat{\text{j}}$
$|\vec{\text{v}}|=\sqrt{(3)^2+(-8)^2}$
$=\sqrt{9+64}=\sqrt{73}\text{ m/s}$

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