Question

A particle moves along a curve given by x = 2t^2, y = t^2 - 41, and 2 = 31-5 where t represents time. Find the velocity and acceleration in the direction X-3y + 22 att=1

Answer 1

Answer with Step-by-step explanation:

We know that the components of velocity are obtained from position as

$u=\frac{dx}{dt}\\\\v=\frac{dy}{dt}$

Using the given values we obtain

$u=\frac{d(2t^2)}{dt}\\\\u=4t$

Similarly

$v=\frac{d(t^2-41)}{dt}\\\\u=2t$

The the velocity function can be written as

$\overrightarrow{v}=4t\widehat{i}+2t\widehat{j}$

The components of acceleration are obatined from the components of velocity as

$a_{x}=\frac{du}{dt}\\\\a_{y}=\frac{dv}{dt}$

Using the given values we obtain

$a_x=\frac{d(4t)}{dt}\\\\a_{x}=4$

Similarly

$a_y=\frac{d(2t)}{dt}\\\\a_y=2$

The the acceleration function can be written as

$\overrightarrow{a}=4\widehat{i}+2\widehat{j}$

Thus at time 't=1' the velocity function becomes

$\overrightarrow{v}=4\widehat{i}+2\widehat{j}$

Thus the component of acceleration in the direction of the given vector $\overrightarrow{r}=\widehat{i}-3\widehat{j}$ can be found by taking the dot product of the 2 vectors

Thus we get

$v_{r}=\overrightarrow{v}\cdot \overrightarrow{r}\\\\v_{r}=(4\widehat{i}+2\widehat{j})\cdot (\widehat{i}-3\widehat{j})\\\\v_{r}=4-6=-2$

Similarly the dot product is obtained for acceleration as

$a_{r}=\overrightarrow{a}\cdot \overrightarrow{r}\\\\a_{r}=(4\widehat{i}+2\widehat{j})\cdot (\widehat{i}-3\widehat{j})\\\\a_{r}=4-6=-2$