Homework help with interpreting SUTAV questions

  • Thread starter Physics_is_beautiful
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  • #1
Physics_is_beautiful
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Homework Statement
The question states :

--> A person and a car are in a race together.

-->The runner runs at a constant speed of 10 m/s from the start of The race. During the first 5.0 s of the race, the car's speed increases from 0m/s to 25 m/s at a uniform rate.

then asks :

--> At what point of time does the car over take the runner?

[Same exact words as per question]
Relevant Equations
all basic equations of motions,
but particularly :

a = (v-u)/t
s = ut+1/2at^2
here's what I did :

I tried to balance both sides of the equations, to find the point where both the car and the runner are at the same place, thus ;

--> ut + 1/2at^2 = ut + 1/2at^2

--> (10)(t)+1/2(0)(t^2) = (0)(5) + 1/2(5)(t^2)

[ Here LHS if for the runner, RHS for the car. I do realise that the speed of a human cannot be 0 right at the start, as in, he needs to accelerate due to inertia, but the question states that he's at 10 m/s from the start, so T = 0.0s is a point in time where the speed of the car = 0 m/s and the the speed of the person in 10m/s]

-->10t = 2.5t^2

--> 4t = t^2

--> t = 4 s

However, the question asks specifically at what point in time does the car overtake the runner, which is Immediately after t = 4.0s.
But, when exactly?

since the time is given to 2 sig-figs, I gave the answer as approx 4.1s, but I am not too sure of that.

That's where I need help.

Thanks.
 
Last edited:
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  • #2
Physics_is_beautiful said:
However, the question asks specifically at what point in time does the car overtake the runner, which is Immediately after t = 4.0s.
But, when exactly?
Same time (4s). No need to play with words (there are so many things that are approximated here, included the size of the runner and the car). Reaching the same point and overtaking mean the same thing here.
 
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  • #3
Physics_is_beautiful said:
a = v-u/t
Did you mean a = (v-u)/t ?
 
  • #4
kuruman said:
Did you mean a = (v-u)/t ?
yup. I'll just edit it.

edit : done.
 
  • #5
DrClaude said:
Same time (4s). No need to play with words (there are so many things that are approximated here, included the size of the runner and the car). Reaching the same point and overtaking mean the same thing here.
Yeah, I didn't think about that either. When a car has reached a point (let's say point 'A'), the same as the runner, it has already kind of overtaken him cause the car would be much longer. However, once could argue if overtaking means that the car is entirely in front of the runner or something like that.

You're right. It's kind of a mess.

But it's a grade 9 question, so I doubt it would require so much of thought. Plus, we had 1 mark for this question.
 
  • #6
Physics_is_beautiful said:
Yeah, I didn't think about that either. When a car has reached a point (let's say point 'A'), the same as the runner, it has already kind of overtaken him cause the car would be much longer. However, once could argue if overtaking means that the car is entirely in front of the runner or something like that.

You're right. It's kind of a mess.

But it's a grade 9 question, so I doubt it would require so much of thought. Plus, we had 1 mark for this question.
Think of it this way. For the race to be fair, a point on the car must be lined up with a point on the runner at the start of the race. "Overtaking" would mean that the two points are lined up again father down the track. This is equivalent to having a "point" car and a "point" runner which is what the SUVAT equations assume. So don't worry about it.
 

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