Answer:
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
