How Far Did Yatin Travel Before Stopping at the Intersection?

AI Thread Summary
Yatin was driving at 90.0 km/h and had to decide whether to stop at a yellow light 75.0 meters ahead, taking 0.400 seconds to make his decision. After recalculating, the stopping distance was determined to be 2.5 meters before the intersection, indicating he would have stopped safely. The calculations involved determining the distance traveled during the decision-making time and the braking distance using the correct formulas. The discussion highlighted the importance of using accurate formulas for distance and time in physics problems. Ultimately, Yatin made the right decision to stop safely at the intersection.
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Stopping Distance Grade 11 Physics Help Needed -- URGENT!

Yatin was driving at 90.0km/h when he saw the light turn yellow at an intersection located 75.0m ahead. Yatin used 0.400s to make up his mind, and then braked at 5.00m/s2 until he stopped. What was his stopping distance? Did Yatin make the right decision? If not, what could he have done differently?



(ATTEMPT BELOW COPIED/PASTED FROM MS WORD EQUATION EDITOR [MY WORK])
v1=90.0km/h or 25.0m/s
a=-5.00/s/s
v2=0.00m/s
Step 1:
t=v1/a
t=25.0/5.00
t=5.00s
Step 2:
d after making decision=d-(v1*t)
d after making decision=75-(25*.04)
d after making decision=75-1
d after making decision=74.0m

Step 3:
stopping distance=dtotal-(a*t)
stopping distance=74m-(5*5)
stopping distance=74-25
stopping distance=49m
Therefore Yatin`s stopping distance is 49m
 
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Hello Newb, welcome to PF!
In step 2, you used a time of .04 s when the question said 0.4 s.
In step 3, you used a*t as a distance, but in fact it is a velocity.
It seems to me the solution would be clearer if you just calculated the distance for each of the two motions, then add them up and see if they exceed the 75 m.
 


Thanks delphi...i think i fixed it:
Yatin was driving at 90.0km/h when he saw the light turn yellow at an intersection located 75.0m ahead. Yatin used 0.400s to make up his mind, and then braked at 5.00m/s2 until he stopped. What was his stopping distance? Did Yatin make the right decision? If not, what could he have done differently?
v1=90.0km/h or 25.0m/s
a=-5.00/s/s
v2=0.00m/s
Step 1:
t=v1/a
t=25.0/5.00
t=5.00s
Step 2:
d after making decision=d-(v1*t)
d after making decision=75-(25*.400)
d after making decision=75-10
d after making decision=65.0m

Step 3:
stopping distance=dtotal-[(a*t)*t]
stopping distance=65m-(5*5)*5
stopping distance=65-25*5
stopping distance=65-125
stopping distance=-60m
Therefore Yatin`s stopping distance is 60m after the light.


Is this right? I didn't understand what you meant by calculating the distance of both motions?
 


Would really appreciate help...The thing is due in a couple hours...i have 2 more questions to do :S...tensed!
 


Step 3 - you used a*t*t for the distance, but this still is not the correct formula for distance. The negative distance means the car has reversed direction - a clear indication that something is wrong in the calculation.
 


These were the only formula's given to us...would you be able to tell me which one to use in this case?
http://img196.imageshack.us/img196/7408/captureie.png
 
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Any of the ones with a D in them should work, if you can figure out the stopping time first.
 


wait so do you mean the stopping time i calculated was wrong?
Step 1:
t=v1/a
t=25.0/5.00
t=5.00s
 


Oh, I forgot you had that. It is correct!
 
  • #10


ok i think i got it now...is this correct?
v1=90.0km/h or 25.0m/s
a=-5.00/s/s
v2=0.00m/s
Step 1:
t=v1/a
t=25.0/5.00
t=5.00s
Step 2:
d after making decision=d-(v1*t)
d after making decision=75-(25*.400)
d after making decision=75-10
d after making decision=65.0m

Step 3:
stopping distance=d- [(v1+v2)/2] t
stopping distance=65-[(25+0)/2]*t
stopping distance=65-(12.5*5)
stopping distance=65-62.5
stopping distance=2.5m
Therefore Yatin`s stopping distance is 2.5m before the light.
 
  • #11


Got it!
Would perhaps be nicer if you wrote
d = vt + [(v1+v2)/2] t
= 25*0.4 + (25 + 0)/2*5
= 72.5 m
 
  • #12


Got it Thanks
 
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