A police dog gets caught in a shootout. A 5 g bullet moving at 100 m·s–1 strikes the dog and lodges in his shoulder. Fortunately the dog survives. The bullet undergoes uniform acceleration and penetrates his shoulder to a depth of 6 cm. Calculate the time taken for the bullet to stop

Since the bullet undergoes uniform acceleration, its average speed while inside the dog is 1/2 of (initial speed + final speed) = 1/2(100 + 0) = 50 m/s .It travels 6 cm at that average speed.Time = (distance) / (speed) = (6 cm) / (50 m/s) = (0.06 m) / (50 m/s) = 0.0012 sec(1.2 milliseconds) . 

 V² = U² + 2 a SHere V = final velocity = 0  of the bullet as it stopped       U = initial velocity before the bullet entered the shoulder = 100 m/s       a  = ?       S = distance traveled by bullet under the acceleration/deceleration. = 6 cm         = 0.06 metera = (0² - 100²) / 2 * 0.06  =  - 83,333 meters/sec²  = - 8.333 * 10^4 m/sec²It is negative as it is a deceleration.      Now ,let us calculate the time duration T taken by the bullet to stop after entering the shoulder.V = u + a t      Here, we have V = final velocity = 0.    u is 100 m/s    a is found above.So   0  = 100 - 83333  T         =>      T =  100 / 83333  = 1.2 milliseconds