A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m
ust stop the car. If it takes 0.200 s for the driver to apply the brakes, what must be the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign? A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she must stop the car. If it takes 0.200 s for the driver to apply the brakes, what must be the magnitude of the constant acceleration of the car after the brakes are applied so that the car will come to rest at the stop sign? 3.89 m/s2 4.17 m/s2 2.89 m/s2 3.42 m/s2 2.08 m/s2
We are told the reaction time is 0.2 s. Now, during this reaction time the car is going to travel an additional distance of
: x = u × t = 40 × 0.2 = 8 m
where u is the initial velocity of the car which is 40.0 m/s.
We are told that he had 200 m to stop before applying brakes. Thus, after applying brakes, he now has a distance to cover of; s = 200 - 8 = 192 m
Since vehicle is coming to rest acceleration would be negative, thus using Newton's equation of motion, we have;
The net charge on an atom is equal to the overall difference between the number of protons in the nucleus versus the number of electrons around the nucleus, where a negative sign represents less protons and a positive sign represents more protons (than electrons).
