A driver in a car traveling at a speed of 78 kilometers/h sees a cat 101 m away on the road . How long will it take for the car to accelerate uniformly to a stop in exactly 99 m?

0.353 secs

Explanation:

Using the formula;

ΔS = ut + 1/2at²

ΔS is the change in distance = 101 - 99 = 2m

u is the speed = 78km/hr = 7800/3600 m/s

Speed = 2.17m/s

a is the acceleration = 9.8m/s262

t is the time

Substitute into the equation and get t

2 = 2.17t + 1/2(9.8)t²

2 = 2.17t + 4.9t²

2.17t + 4.9t²- 2 = 0

Factorize

t = -4.9±√4.9²-4(2.17)(-2)/2(2.17)

t = -4.9±√24.01+17.36/4.34

t = -4.9±6.43/4.34

t = -4.9+6.43/4.34

t =1.53/4.34

t = 0.353secs

Hence it will take the car 0.353 secs

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge