Question

He reaction time for an automobile driver is 0.7 sec. if the automobile can be decelerated at 5km/s .calculate the total distance travelled in coming to stop from an intial velocity of 8.33 m/s after a signal is observed.

Answer 1

Find the distance covered before the driver reacts and the distance covered during the deceleration after he reacts.
1)The distance covered before he reacts:
=reaction time x speed
5km/hr= 1.389 m/s
= 8.33 x 0.7
=  5.831m

2)distance covered during deceleration:
use kinematic equations:
SUVAT-- s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-time
v=u + at
v2 = u2 + 2as
s=1/2(u+v)t
s=ut + 1/2at2

----->  v=0, u=8.33,a=-1.389 (negative acceleration)
-----> v2=u2+2as
         0=8.33^2 + 2 x -1.389 x S
         0=69.44 - 2.778S
         2.778s= 69.444
                 s= 69.44/2.778
                   =24.998m

Calculate total distance covered=  24.998 + 5.831
                                              = 30.83m

Therefore distance covered = 30.83 meters

Answer 2

Explanation:

hope you have understood......................