Question

If the speed of a truck is reduced from 26.7m/s to 6.7m/s within a distance of 800m. Find to How long were the breaks applied? How much longer would it take before coming to rest?​

Answer 1

Below

Step-by-step explanation:

AVERAGE speed of the truck is ( 26.7 + 6.7) /2 = 16.7 m/s

the time required at this avg speed to cover 800 m is

800 m / 16.7 m/s = 47.9 s

acceleration is change in velocity / change in time

( 6.7 - 26.7) / 47.9 = - .4175 m/s^2

Now: vf = vo + at

0 = 6.7 + (-.4175)t shows t = 16 more seconds to stop