Question

Use the position function s(t) = –4.9t2 + 500, which gives the height (in meters) of an object that has fallen for t seconds from a height of 500 meters. The velocity at time t = a seconds is given by the following.
lim t->a: (s(a) - s(t)) / (a - t)
Find the velocity of the object when t = 3. (Round your answer to two decimal places.)

Answer 1

Complete Question

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Answer:

The  value is  $v(3) = 29.4 \ m/s$

Step-by-step explanation:

From the question we are told that

    $s(t) = -4.9t^2 + 500$

And  

    $\lim_{t \to a} \frac{s(a) - s(t)}{a-t}$

Generally s(t) at t = a is mathematically evaluated as

      $s(a) = -4.9a^2+ 500$

So

    $s(a) - s(t) = -4.9a^2 + 500 - ( -4.9t^2+500)$

     $s(a) - s(t) = -4.9a^2 + 500 +4.9t^2-500)$

      $s(a) - s(t) = 4.9(t^2 - a^2)$

Thus the velocity is represented as

     $v(t) =\lim_{t \to a} \frac{s(a) - s(t)}{a-t} \equiv \lim_{t \to a} \frac{4.9(t^2 - a^2 )}{ a-t}$

=>     $v(t) = \lim_{t \to a} - 4.9(a + t )$

=>      $v(t) = -4.9 (a + a )$

=>      $v(t) = -9.8a$

Now at  t =  3

=>  $v(3) = -9.8 (3)$

=>   $v(3) = 29.4 \ m/s$