Question

The instantaneous coordinates of a particle are x=(4t-1) m and y=8t^2 m. Calculate average velocity at time interval 1s to 2s

Answer 1

The average velocity of the particle is
`\vec {v}_(avg) = 4\,\hat{i} +24\,\hat{j}\,\left[(m)/(s) \right]`.

Explanation:

In vector terms, the average velocity of a particle (
`\vec v_(avg)`), in meters per second, at a given time change (
`\Delta t`), in seconds:

`\vec v_(med) = (\Delta x)/(\Delta t)\,\hat{i} + (\Delta y)/(\Delta t)\,\hat{j}` (1)

Where
`\Delta x` and
`\Delta y` is the change in position for x and y axes, in meters.

`\Delta x = x(2) - x(1)`

`\Delta x = (4\cdot 2 - 1)-(4\cdot 1 - 1)`

`\Delta x = 4\,m`

`\Delta y = y(2) - y(1)`

`\Delta y = 8\cdot (2)^(2)-8\cdot (1)^(2)`

`\Delta y = 24\,m`

`\Delta t = 2\,s - 1\,s`

`\Delta t = 1\,s`

`\vec v_(avg) = \left((4\,m)/(1\,s)\right)\,\hat{i} + \left((24\,m)/(1\,s) \right) \,\hat{j}`

`\vec {v}_(avg) = 4\,\hat{i} +24\,\hat{j}\,\left[(m)/(s) \right]`