Question

Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time t, find the body's position at time t. a=16, v(0) = -14, s(0) = -8

Answer 1

Answer:

The  position at time t  is  $s(t) = 8t^2 - 14t -8$

Step-by-step explanation:

From the question we are told that

   The acceleration is  $a = 16$

   The velocity at t = 0  is  $v(0) = -14$

    The  position at time t =  0  is $s(0) = -8$

Generally acceleration is mathematically represented as

     $a(t) = \frac{d v}{dt }$

=>   $\frac{dv}{dt} = 16$

=>   $dv = 16 dt$

integrating both sides we have    

     $\int\limits dv = \int\limits16 dt$

=>  $v(t) = 16 t + c$

Now at t = 0  

     $v(0) = 16 * 0 +c = -14$

=>  $c = -14$

So

   $v(t) = 16 t -14$

Generally the position of the body is mathematically represented as

     $s(t) = \int\limits v(t)dt$

So

    $s(t) = \int\limits 16t - 14 dt$

So  

    $s(t) = 16 \frac{t^2}{2} - 14t + C$

Now  at t =  0

     $s(0) = 16 \frac{0^2}{2} - 14(0) + C = -8$

=>    $C = -8$

So

   $s(t) = 8t^2 - 14t -8$