When an event occurs that requires an emergency stop, the vehicle continues to t

Question

When an event occurs that requires an emergency stop, the vehicle continues to travel at its initial velocity while the driver reacts to the event. The distance traveled for this time is the perception-reaction distance. Secondly, the vehicle undergoes constant acceleration while the brakes are applied. The distance traveled for this time interval is the braking distance. Historically, engineers have used a perception-reaction time of 0.75 seconds, but they now assume a perception-reaction time of 1.0 seconds for the average driver. A vehicle has an initial velocity of v_0 when a tree falls on the roadway a distance x_f in front of the vehicle. The driver has a perception-reaction time of t. When the driver begins braking, the vehicle has an acceleration of magnitude a. write an expression for the velocity of the vehicle when it hits the tree. A vehicle has an initial velocity of V_0 = 31ms when a tree falls on the roadway x_f = 84 m in front of the vehicle. When the driver begins braking, the vehicle has an acceleration of magnitude a = 6.4 m/s^2. The driver has a perception-reaction time of 0.75 s. Calculate the velocity of the vehicle when it hits the tree. A vehicle has an initial velocity of V_0 = 31 m/s when a tree falls on the roadway x_f = 84 m in front of the vehicle. When the driver begins braking. the vehicle has an acceleration of magnitude a = 6.4m/s^2. The driver has a perception-reaction time of 1.00 s. Calculate the velocity of the vehicle when it bits the tree,

Explanation / Answer

It is unclear if you are asking for all the answers or just first. I will answer only the first one & request you to upload others seperately.

s = s1 + s2, where s1 is the distance travelled during perception reaction & s2 is distance travelled during decelaration

s1 = v0t

For s2, we use third equation of motion,
v2-u2 = 2as
0-v02=2*(-a)*s2

s2 = v02/2a
s = v0t + v02/2a

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