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=(dx)/(dt)\\\\v=(dy)/(dt)`

Using the given values we obtain

`u=(d(2t^2))/(dt)\\\\u=4t`

Similarly

`v=(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)=(du)/(dt)\\\\a_(y)=(dv)/(dt)`

Using the given values we obtain

`a_x=(d(4t))/(dt)\\\\a_(x)=4`

Similarly

`a_y=(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`