Question

the position from its starting point of a small plane preparing for takeoff is given by x(t)= 1.43t^2 meters…

Answer 1

From the given problem, the position of the plane is at :

`x(t)=1.43t^2`

First step is to determine the velocity.

Velocity is the 1st derivative of the position.

Note that the general differentiation is :

`d(ax^n)=n(ax^(n-1))`

The velocity will be :

`\begin{gathered} V(t)=dx(t) \\ V\mleft(t\mright)=2\mleft(1.43t\mright) \\ V(t)=2.86t \end{gathered}`

Acceleration is the 1st derivative of the velocity.

So it follows that :

`\begin{gathered} a(t)=dV(t) \\ a(t)=1(2.86)\text{ } \\ a(t)=2.86 \end{gathered}`

Therefore, the answer is 2.86 m/s^2