Question

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop

Answer 1

Answer:

The correct solution is "122.2211".

Explanation:

Given:

deceleration,

a = 22 ft/sec²

Initial velocity,

$V_i=50 \ m/h$

Now,

$V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s$

    $=73.333 \ ft/sec$

Now,

Final velocity,

$V_f=0$

Initial velocity,

$V_{initial} = 73.333 \ ft/sec$

hence,

⇒ $V_f^2=V_i^2+2aD$

By putting the values, we get

      $0=(73.333)^2+2\times( -22) D$

  $44D=(73.333)^2$

      $D=\frac{(73.333)^2}{44}$

          $=122.2211$