Question

Suppose f (t) = −11t2 is the position at time t of an object moving along the x axis. Use the limit definition to find the velocity of the object at time t0 = −5.

Answer 1

The velocity of the object at time t0 = -5 can be found by taking the derivative of the position function, which results in v(t) = -22t. Substituting t = -5 into this velocity function yields v(-5) = 110 m/s.

Step-by-step explanation:

The student asks for the velocity of an object at time t0 = -5 given the position function f(t) = −11t2. To find the velocity, we use the limit definition of the derivative. The derivative of the position function with respect to time gives the velocity function:

v(t) = f'(t) = d/dt (-11t2)

This results in:

v(t) = -22t

Now, we can find the velocity at t0 = -5 by plugging the value into the velocity function:

v(-5) = -22(-5) = 110 m/s

Therefore, the velocity of the object at t0 = -5 is 110 m/s.

Answer 2

The velocity is
`v(-5)=110`.

Step-by-step explanation:

We have the position as a function of time:
`f (t) = -11t^(2)`

Instantaneous velocity at
`t=t_(0)` is defined as:

`v(t_(0))= \lim_(\triangle t \to 0) (f(t_(0)+\triangle t)-f(t_(0)) )/(\triangle t)`

If
`t_(0)=-5`,

`v(-5)= \lim_(\triangle t \to 0) (f(-5+\triangle t)-f(-5) )/(\triangle t)=\lim_(\triangle t \to 0) (11(-5+\triangle t)^(2) -11(-5)^(2)  )/(\triangle t)`

`=\lim_(\triangle t \to 0) (11(25+10\triangle t+\triangle t^(2))  -275  )/(\triangle t)=\lim_(\triangle t \to 0) (110\triangle t+11\triangle t^(2))/(\triangle t)`

`\lim_(\triangle t \to 0) ((110+11\triangle t)\triangle t)/(\triangle t)=\lim_(\triangle t \to 0) 110+11\triangle t=110`

So,

`v(-5)=110`.