Question

Problem-1 Nardine is traveling southbound on a level section of the 15 freeway with speed up = 60mi/hr. Joshua is also traveling southbound a distanced directly behind Nardine with the same speed to 60mi/hr. Nardine has ABS brakes in her car, and Joshua does not have ABS brakes in his car. On this section of the freeway, Nardine's tires have coefficients of friction (M ) (0.7,0.5) and Joshua's tires have coefficients of friction of Walk) = (0.5, 0.3). Nardine suddenly notices something blocking the road in front of her and slams on her brakes. Suppose that Joshua's reaction time is tr=0.250s. What is the minimum safe distance dmin that the front of Joshua's car needs to be from the back of Nardine's car in order to avoid Joshua crashing into Nardine?

Answer 1

initial vel v_o = 60 mi/hr = 88ft/s

when breaks are applied the car will decelerate with the frictional force.

frictional force F = -_$\mu$kmg

deceleration a= F/m =  -_$\mu$kmg

deceleration for Nardine car a_n = -0.5*9.8 = -4.9 m/s/s

time to stop v_o  - 4.9*t =0

t_n = 88/4.9 = 17.96 s

distance moved v_o = 2as

s_n = v_o²/2a = 790.2 ft

deceleration for Joshua car a_j  = -0.3 *9.8 = -2.94 m

time to stop t_j = 88/2.94 = 29.93 s

distance moved s_j = 88²/2*2.94 = 1317 ft

response time = 0.25 s

for 0.25 s Joshua moves with speed v_o = 60 mi/hr = 88ft/s

distance moved = 88*0.25 = 22ft

Nardine will  stop in 790.2 ft and

Joshua will move 1317 +22 = 1339 ft before coming to rest.

safe distance to avoid collision is 1339 - 790.2 = 527 ft