Helpp homework ( I know there's a section for it, but I don't understand how to post)

  • Thread starter MeghanReid
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  • #1
1. A man is driving his new sports car at 120km/h while a girl is parked at a red light. At the moment the man passes the girl, the light turns green and the girl pulls away at the same time as the man passes. The girl accelerates at 45km/h.

A) How long will it take the girl to catch the man?
B) How far will the girl have to travel to catch the man?

2. A women is 0.056 km ahead of a man in a race. The women is jogging at a constant rate of 3 m/s while the man is jogging 2.3 m/s until he decides to catch up to the women and accelerates at 1.2 m/s.

A) How far does the man have to travel to catch up to the women?

Please help. I don't understand these at all.

Answers and Replies

  • #2
Simon Bridge
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The girl accelerates at 45km/h
That is not an acceleration.

The way to do this is draw the speed-time graph.
Since the girl starts at rest, the graph will be a triangle.
The slope of the line is the acceleration. a=v/T (rise over run)
The area under the graph is the distance. d=vT/2 (area of a triangle)

same with the next one - draw the graph. This time it is not a triangle because the man starts at speed.
  • #3

I can't draw a speed time graph because I don't know the time. That's part of what I have to find in the question. I know I have to use the formulas to solve it and "compare" the two people and stuff, but I dont know how to do so.
  • #4
Simon Bridge
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Mark a point on the t axis and label it T.
When you need to put a time in the equations for slope and area, just write T.

say the girl starts out at v=0, and end up at v=120kmph, and does this in some time T.
so on the graph, mark off 120 on v axis and just pick a spot for T on the t axis.
draw a line from (0,0) to (T,120).

now you have a triangle with one unknown side.

the acceleration is the slope of the line
45=120/T (rise over run get it?) so no-brainer: T=120/45

(still need the proper units for that acceleration - 45kmph/s is quite impressive. 0-100 in just over 2 seconds.)

the distance is the area under the graph - half the base times the height because it is a triangle:

d=120T/2 ... and you've just worked out T :)

the beauty of this method is you have no confusing kinematic equations to memorize.

Of course - for the girl to catch the man, you need the time where the displacements are equal.
This means the girl can't just stop at 120 - she's still behind.

The trick is to find the T so that the mans distance and the girls distance are the same.
So you need the mans v-t graph too.

The girl is still:

you don't know the final speed - but you do know:
45=v/T which is v=45T so

dg = (12.5)T^2

the man's graph is just a rectangle of height 120 and length T (same T as the girl).

since his line is flat:

the two distances are the same when dg=dm => (12.5)T^2 = 120T
which is a quadratic equation. Solve for T.

(there's a shortcut - vg=2vm ... can you see why?)
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  • #5

Okay thank you. I figured it out (:
  • #6

I'm having trouble with number 2 though.. The graph doesn't help me much, and I've been taught by my teacher to use formulas not graphs to solve word problems. For example:


are the 2 main formulas we use pretty much.
  • #7
Simon Bridge
Science Advisor
Homework Helper

... but then you have to memorize the formulas, and try to avoid getting confused about what goes where. Also you end up concentrating so much on the formulas that you don't see the physics. It's the physics that's important. (Those are the right equations btw - bet you cannot figure out how to apply them to #2! - well, I know you can't because you are asking me!)

Note: you still have to fix the units for acceleration - you will lose marks if you don't.

This time your graph is much the same except it is the woman who is constant and the man who changes ... and his line does not start from v=0.

the man and the woman travel at constant speed until the woman is ahead of the man by a given amount. So draw that in. You know the two speeds, but you don't know the time so put a T1 in on the time axis (because there are two parts to the motion).

The the man accelerates until time T2, when he's caught up again.

We want the distance traveled by the man between T1 and T2.
As before, the total distance traveled (from 0 to T2) by both of them is equal.
You just have a more complicated area - that's all.
You deal with complex areas by dividing them into simple areas.
The man's graph is a rectangle and a trapezium.
  • #8
Simon Bridge
Science Advisor
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That's what I said :)

There's a shortcut with the graph too but you are having a hard enough time understanding what is going on without me showing off my math-foo. I'll stick to the standard method:

Here's a more general form of the diagram...

u is the speed of the woman jogging. the man starts out at v until he decides to put a spurt on at t=T1. Can you see that the lead the woman has by T1 is the area between u and v in this region? (distance is area remember?)

Since you know u and v you can get T1. That leaves T2 and w you don't know.

Since this is homework, there is a limit to how much I can tell you.
Go through the geometry and leave the variable every time you don't know something.
Use algebra to find the missing values.


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  • #9
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MeghanReid said:
Helpp homework ( I know there's a section for it, but I don't understand how to p
Hello Meghan, welcome to Physics Forums.

In the future, please post physics homework questions here:

I'll go ahead and move this thread there.

Also, please include the equations that are relevant to the problem whenever you post a homework problem -- and show how far you have gotten with the problem.

... but then you have to memorize the formulas, and try to avoid getting confused about what goes where.
Maybe, but a lot of teachers either supply the formulas or let students bring a notecard or sheet of formulas to exams. Though I agree that drawing a graph is an excellent way to understand what is going on.

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