Let u = the speed of the car at the instant when braking begins.
The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.
Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
u = 27.2546 m/s
Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
= 2.7696 m/s
Answer: 2.77 m/s (nearest hundredth)
The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have

where C are the concentration values and t is the time taken for it to decompose.


Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s
b is your answers in this thread