Question

If the position of aparticle is given by x=5.0-9.8t+6.4t^2, what is the velocity and acceleration of the particle at t=4.0s?

Answer 1

`v(4) = 41.4 \text( m/s)`
`a(4) = 12.8 \text( m/s)^2`

Explanation:

`x(t) = 5.0 - 9.8t + 6.4t^2 \text( m)`

`v(t) = (dx(t))/(dt) = -9.8 + 12.8t\text( m/s)`

`a(t) = (dv(t))/(dt) = 12.8\text( m/s)^2`

At `t = 4`:

`v(4) = -9.8+12.8(4) = 41.4 \text( m/s)`
`a(4) = 12.8 \text( m/s)^2`

Answer 2

The given equation can be compared with `s=ut +1/2 at^2`

which is an equation of position-time relationship of a particle moving with constant acceleration.

So,rearranging the given equation,we get,

`x=5-9.8*t +1/2 *12.8 t^2` (also see, at `t=0,x=5`)

So,the acceleration of the particle is constant i.e `12.8 ms^-2` and initial velocity `u=-9.8 ms^-1`

Now,we can use the equation, `v=u +at` to find velocity after `4s`

So, `v=-9.8 +12.8*4=41.4 ms^-1`