The answer to the question is True
As it is given that the air bag deploy in time
total distance moved by the front face of the bag
Now we will use kinematics to find the acceleration
now as we know that
so we have
so the acceleration is 400g for the front surface of balloon
It would be 20 protons left because 1-19= 20
Your position in meters will, measured relative to the starting point of the car behind you, be
x1(t) = 10 + 23.61 t - 1/2 4.2 t^2
his position will be
x2(t) = 16.67 t
Hence at any time the separation s(t) will be
s(t) = x1(t) - x2(t) = 10 + 6.94 t -2.1 t^2
Now I assume you mean that you will decelerate UNTIl you are driving at the legal speed limit (60 km/h). That will take you:
16.67 m/s = 23.61m/s - 4.2 m/s^2 * t
t = 1.65 seconds
What is the separation at that time? If it is still greater than zero, there will be no collision:
s(1.65) = 10 + 6.94 *1.65 -2.1 (1.65)^2 = 15.73 meters.
Hence you will NOT collide. The 1.65 s you calculated was the time needed to brake to the speed of 60 km/h.
