This is too hard for me. Someone help me~ *Average Reaction time in Part 1 Secti

Question

This is too hard for me. Someone help me~

*Average Reaction time in Part 1 Section E is 155ms ..... this will help with #3

E Applied Problem: Consider the case where a person needs to stop a car to avoid hitting an object in the road. In order to stop, all drivers must first react and then, apply the brakes. Thus, to determine the total distance needed to stop one must account for two distances: 1. the distance traveled during the reaction time before the brakes are applied (i.e. the car is traveling at constant speed). 2. the distance traveled during the actual braking (i.e. the car accelerates).

Explanation / Answer

3. Distance travelled during reaction time:

As no acceleration is involved, d = v*t

for v = 16 m/s; d = 16* 155 = 2480 m

for v = 25 m/s; d = 25*155 = 3875 m

for v = 34 m/s; d= 34*155 = 5270 m

4. http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Kinematics/BrakingDistData.html -- Source on deceleration rates for various models.

5. Using deceleration rate of 7.5 m/s^2

From the moment brakes are applied, distance travelled can be computed as:

v^2 - u^2 = 2*a*s

V is final velocity, which will be zero as car is coming to a halt, u is initial velocity, a is -7.5 m/s^2

for u= 16 m/s, 16^2 = 2*7.5 *s => s = 17.06

for u = 25 m/s, 25^2 = 2*7.5*s => s = 41.67

for u = 34 m//s, 34^2 = 2*7.5*s => s = 77.06

total distance travelled:

for u = 16m/s => 2480 + 17.06 = 2497.06 m

for u = 25 m/s => 3875 + 41.67 = 3916.67 m

for u = 34 m/s => 5270 + 77.06 = 5347.06 m

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