How Far Did Yatin Travel Before Stopping at the Intersection?

In summary, Yatin was driving at a speed of 90.0km/h when he saw the light turn yellow at an intersection 75.0m ahead. After taking 0.400s to make a decision, he braked at a rate of 5.00m/s2 until he stopped. His total stopping distance was 49m and his stopping distance before the light was 2.5m. Yatin could have made a better decision by starting to brake earlier or at a slower rate.
  • #1
physics_newb
40
0
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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  • #2


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.
 
  • #3


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?
 
  • #4


Would really appreciate help...The thing is due in a couple hours...i have 2 more questions to do :S...tensed!
 
  • #5


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.
 
  • #6


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
 
Last edited by a moderator:
  • #7


Any of the ones with a D in them should work, if you can figure out the stopping time first.
 
  • #8


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
 
  • #9


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
 

Related to How Far Did Yatin Travel Before Stopping at the Intersection?

What is stopping distance?

Stopping distance is the total distance a moving object travels from the moment the brakes are applied until it comes to a complete stop.

What factors affect stopping distance?

The main factors that affect stopping distance are the speed of the object, the mass of the object, and the efficiency of the brakes.

How is stopping distance calculated?

Stopping distance can be calculated using the formula: d = v2 / 2a, where d is the stopping distance, v is the initial velocity, and a is the deceleration.

How does speed affect stopping distance?

The higher the speed of the object, the longer the stopping distance will be. This is because the object has more kinetic energy and requires more distance to be brought to a stop.

What is the relationship between mass and stopping distance?

The greater the mass of the object, the longer the stopping distance will be. This is because a heavier object has more inertia and requires more force to be stopped.

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