Question

A proton moves along the x axis according to the equation x = 88 t + 11 t^2, where x is in meters and t is in seconds ?

Answer 1

(a). `121"m/s"`

(b). `154"m/s"`

(c). `22"m""/"s""^2"`

Explanation:

`x=88t+11t^2`

(a).

To find the displacement after 3s just substitute this for `t` `rArr`

`x=(88xx3)+(11xx3^2)=363"m"`

The average velocity over 3s = distance travelled/time taken

`=363/3=121"m/s"`

(b).

The instantaneous velocity `V` at a particular time taken `=(dx)/(dt)`:

`(dx)/(dt)=V=88+22t`

If `t=3` this becomes:

`V=88+(22xx3)=154"m/s"`

(c).

Acceleration `a` is the rate of change of `V` with time. So we differentiate the expression for `V`:

`a=(dV)/(dt)=22"m""/"s""^2"`