Question

CarA starts from rest at t= 0 and travels along a straight road with a constant acceleration of 6 ft/s^2 until it reaches a speed of 80 ft/s . Afterwards it maintainsthis speed. Also, when t= 0, carB located 6000 ft down the road is traveling towards at a constant speed of 60 ft/s .

Determine the distance traveled by car A when they pass each other?

Answer 1

Car A starts from rest at t=0 and travels along a straight roadwith a constant acceleration of 6ft/s^2 until it reaches aspeed of 80ft/s. After wards it maintains this speed. Also, whent=0, car B located 6000 ft down the road is traveling towards A ata constant speed of 60ft/s. Determine the distances traveled by carA when they pass each other.

for car A
initial velocity u=0m/sec
acceleration=a=6ft/sec2
speed attained by A =80 feet/sec
time taken by A to attain the speed =t
t=[v-u]/a=[80-0]/6=13.33ft/sec2

distance traveled by car A in this time =s=ut+1/2at2
s= 0+0.5*6*80/6*80/6=533.33feet
in this time distance traveled by B towards A=80*60/6=800feet
so distance remaining between A and B after 80/6sec=6000-[533.33+800]=4666.67feet.
relative speed between A and B=80-[-60]=140ft/sec
as they are moving in opposite direction.
now time taken by A to cross B =4666.67/140=33.33sec
in this time distance traveled by A=80*33.33=2666.66feet.
hence the distances traveled by car A when they pass eachother=533.33+2666.66=3199.99=3200feet.

Answer 2

3200ft.

Answer 3

Car A starts from rest at t=0 and travels along a straight roadwith a constant acceleration of 6ft/s^2 until it reaches aspeed of 80ft/s. After wards it maintains this speed. Also, whent=0, car B located 6000 ft down the road is traveling towards A ata constant speed of 60ft/s. Determine the distances traveled by carA when they pass each other.

for car A
initial velocity u=0m/sec
acceleration=a=6ft/sec2
speed attained by A =80 feet/sec
time taken by A to attain the speed =t
t=[v-u]/a=[80-0]/6=13.33ft/sec2

distance traveled by car A in this time =s=ut+1/2at2
s= 0+0.5*6*80/6*80/6=533.33feet
in this time distance traveled by B towards A=80*60/6=800feet
so distance remaining between A and B after 80/6sec=6000-[533.33+800]=4666.67feet.
relative speed between A and B=80-[-60]=140ft/sec
as they are moving in opposite direction.
now time taken by A to cross B =4666.67/140=33.33sec
in this time distance traveled by A=80*33.33=2666.66feet.
hence the distances traveled by car A when they pass eachother=533.33+2666.66=3199.99=3200feet.

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