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

The force is
`F =  1164.6\  lbf`

The time is
`\Delta t =  2.44 \  s`

Step-by-step 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 =  (15 *  3600^2)/( 5280) =  36818.2 \  mi/h^2`

Generally the acceleration is mathematically represented as

`a =  (v-u)/(\Delta t)`

=>
`36818.2 =  (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 \ (lbm *  ft)/(s^2)`

Now converting to foot-pound-second we have

`F =  (37500)/(32.2)`

=>
`F =  1164.6\  lbf`