Question

What is its velocity? What is its acceleration?

Answer 1

`x(2.5)=3.435`
`v(2.5)=-1.91`
`a(t)=-2`

Explanation:

It is given that the displacement `x` at time `t` is given by `x(t)=1.96+3.09t-t^2`.

At `t=2.5`, the displacement `x(2.5)=1.96+3.09*2.5-2.5^2=3.435`.

The velocity `v` is the derivative of displacement `x` with respect to time `t`, or `v(t)=dx/dt=3.09-2t`. Thus, the velocity at time `t=2.5` is `v(2.5)=3.09-2*2.5=-1.91`.

The acceleration `a` is the derivative of velocity `v` with respect to time `t`, or `a(t)=(dv)/dt=-2`. Since it stays constant regardless of the time, the acceleration at time `t=2.5` is `-2`.

Answer 2

The particle moves according to the equation, `1.96 + 3.09t - t^2`
so,
`x=1.96 + 3.09t - t^2`
Velocity is the rate of change of displacement, `x`
therefore, `v=dx/dt`

`v=(d(1.96 + 3.09t - t^2))/dt`

`v=(3.09-2t)m/s`

Now, acceleration is the rate of change of velocity, `v`
therefore, `a=dv/dt`

`a=(d(3.09-2t))/dt`

`a=-2` `m/s^2`

Answer 3

Given, `x=1.96+3.09t-t^2`

at, `t=2.50` ,`s=1.96+(3.09*2.50)-(2.50)^2=3.435 m`

differentiating the equation w.r.t time,we get,

velocity =`(dx)/(dt) = 3.09 -2t`

So,at `t=2.50s` its velocity is `3.09 -(2*2.50)=-1.91 m/s`

And,differentiating it one again we get, acceleration= `(dv)/(dt) = -2 m/s^2`

ALTERNATIVELY

You can compare the equation with, `s=ut -1/2 at^2`

So,you can rearrange as, `x-1.96=3.09t - 1/2 ×2×t^2`

So comparing we can say,it had an initial velocity of `3.09 m/s`(at `t=0`) acceleration is `-2 m/s^2` and at `t=0` it was `1.96 m` to the left of origin i.e along negative `X` axis.