Physics: when do use v=d/t versus d=(1/2)(vi+vf)t

[ibeatupnerds]

Question comes from NOVA physics book - Chapter 4, Passage 1, question 1:

...a student accelerates uniformly from rest at one side of the building to the jumping edge, a distance of 5 meters. Just after his feet leave the building, he is traveling horizontally at 5 m/s.
How much time does it take the student to accelerate as he is running along the roof?

Why can't we use Velocity = distance / time. He starts at rest (0 m/s), final velocity is 5 m/s. Distance is 5 m. Get 1 second using this.

According to the books answer:
distance = (1/2)(velocity initial + velocity final)(time). time = 2 seconds


Do I have the velocity = distance/time equation wrong? is velocity average velocity or change in velocity? or did i pick the wrong equation (assuming it's delta velocity and not Velocity average), and in which situation would i use v=d/t versus d=(1/2)(v0+vf)(t)

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[PingPongPro]

Trying to memorize which equations applies to which problems isn't going to be enough for the kinematics portion. It seems that you haven't completely grasped the concepts of acceleration and velocities. Try to think about each problem and what is going on physically, rather than thinking about the equations.

To your question:
You cannot simply use v=d/t here because the velocity is not constant. The question says he is uniformly accelerating, so the velocity at the end (5) cant be used to solve v=d/t. The second equation is essentially the same thing as v=d/t. Since velocity is not constant, we find the avg velocity by doing 1/2(vf+vi) and that Vavg=dt.

Say im driving, and accelerate after a red light uniformly, and reach 30 mph in 2 seconds. You can't simply do 30x2 (velocity times time) to get the total distance. I was not going 30mph the whole time, only by the end.This is why you cant use V=5 in your question, cuz he wasn't going that fast the whole time.

 

[MShopes]

Trying to memorize which equations applies to which problems isn't going to be enough for the kinematics portion. It seems that you haven't completely grasped the concepts of acceleration and velocities. Try to think about each problem and what is going on physically, rather than thinking about the equations.

To your question:
You cannot simply use v=d/t here because the velocity is not constant. The question says he is uniformly accelerating, so the velocity at the end (5) cant be used to solve v=d/t. The second equation is essentially the same thing as v=d/t. Since velocity is not constant, we find the avg velocity by doing 1/2(vf+vi) and that Vavg=dt.

Say im driving, and accelerate after a red light uniformly, and reach 30 mph in 2 seconds. You can't simply do 30x2 (velocity times time) to get the total distance. I was not going 30mph the whole time, only by the end.This is why you cant use V=5 in your question, cuz he wasn't going that fast the whole time.

Well said. In short, when you see two velocities of an object or a person in question, you can't use one velocity. You have to get the average velocity of these two velocities by doing (final velocity+initial velocity)/2 or (0+5)/2=2.5 m/s. That is the velocity you would use in your normal equation (V=d/t). And in fact, without using the longer equation you mentioned, you can use the equation V=d/t or T=d/v to solve your question. Time is now T=5/2.5=2 s. Now you see where the 2 seconds come from?

 

[MShopes]

And you use only V like shown in the question without getting the average velocity when you are sure there is only one velocity involved in the question. There might be one number for velocity but then it might say in words that it for example started from rest, and that is when you have another velocity (initial velocity=0). If there are two velocities, still use V=d/t but make sure to use the average velocity.