Question

(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force and total time required?

Answer 1

Answer:

The  force is  $F = 1164.6\ lbf$

The time is   $\Delta t = 2.44 \ s$

Explanation:

From the question we are told that

  The  mass of the car is  $m = 2500 \ lbm$

   The  initial velocity of the car is $u = 25 \ mi/hr$

   The final  velocity of the car is  $v = 50 \ mi/hr$

  The acceleration is  $a = 15 ft/s^2 = \frac{15 * 3600^2}{ 5280} = 36818.2 \ mi/h^2$

   

Generally the acceleration is mathematically represented as

      $a = \frac{v-u}{\Delta t}$

=>   $36818.2 = \frac{50 - 25 }{ \Delta t}$

=>   $t = 0.000679 \ hr$

converting to seconds

       $\Delta t = 0.0000679 * 3600$

=>     $\Delta t = 2.44 \ s$

Generally the force is mathematically represented as

        $F = m * a$

=>      $F = 2500 * 15$

=>      $F = 37500 \ \frac{lbm * ft}{s^2}$

Now converting to foot-pound-second we have  

         $F = \frac{37500}{32.2}$

=>        $F = 1164.6\ lbf$