Question

A particle moves along X direction with constant acceleration. it velocity after 4s is 18 m/s and displacement is 56m find out its initial velocity and acceleration?

Answer 1

Initial velocity: `10"m"/"s"`

Acceleration: `2"m"/("s"^2)`

Explanation:

We're asked to find the initial velocity `v_(0x)` and the acceleration `a_x` of an particle moving in one dimension with a known velocity and position at a certain time.

To find the initial velocity, we can use the equation

`x = x_0 + ((v_x + v_(0x))/2)t`

We'll take the initial position `x_0` to be `0`. Let's rearrange this equation to solve for the initial velocity `v_(0x)`:

`x/t = (v_x + v_(0x))/2`

`v_(0x) = (2x)/t - v_x`

Plugging in known values, we have

`v_(0x) = (2(56"m"))/(4"s") - 18"m"/"s" = color(red)(10"m"/"s"`

The initial velocity is thus `10"m"/"s"`.

`-------------------`

Now that we know the initial velocity, we can use the equation

`v_x = v_(0x) + a_xt`

to find its acceleration. Rearranging the equation to solve for `a_x`, and plugging in known values, we have

`a_x = (v_x - v_(0x))/t = (18"m"/"s" - 10"m"/"s")/(4"s") = color(blue)(2"m"/("s"^2)`

The acceleration of the particle is thus constant at `2"m"/("s"^2)`.