Horizontal Kinematics - Distance question?

Question

Horizontal Kinematics - Distance question?

Two ants race across a table 62 cm long. One

travels at 5.52 cm/s and the other at 2 cm/s.

When the first one crosses the finish line,

how far behind is the second one? You must answer in m/s. Sorry, I didn't understand the explanation that was posted the first time!

Answer 1

The answer must be in distance (metres or cm) or time (seconds), not velocity

The second ant's velocity is constant at 2 cm/s

it's velocity in m/s = 0.02m/2

time for 2nd ant to finish  

62/2 seconds

= 31 seconds

time for 1st ant to finish

62/5.52

=11.2318840579 seconds

time for second ant to finish after first ant wins

= 31 - 11.2318840579

=19.7681159421 seconds (after 1st ant)

Distance to finish line

= 2 * 19.7681159421

=39.5362318842 cm

= 0.395362318842 m

It's a good thing I have a calculator

You can give your answer to 3 decimal places if you like but it's not accurate

  

Answer 2

I'll do all calculations in cm/s then convert final answer to m/s.

Take a pen and paper and draw a straight line horizontally. Mark the end points as A and B. Let A is the starting point and B is the finish point then AB= 62 cm. Do this step so you can visualize what's going on its easy once you visualize.

Both will start from A obviously the first ant to cross the line would be the one moving at 5.52cm/s as (5.52 cm/s is greater than 2cm/s)

The first ant(one that crosses finish line first is first ant) suppose requires time t to cover 62 cm then

using distance= velocity * time

62=5.52 * t ( Find t this is your first answer i.e at t seconds the first one crosses the finish line)

Obviously since second one hasn't crossed the line yet so it must be somewhere between the points A and B you drew. You can find out how far from point A it is at time t( i.e the time when the first ant has reached B) Say it has reached a point C between Point A and Point B

Again using distance= velocity * t

distance is AC (since start point was A and it has reached C only)

AC=2 * t (put the value of t from the first part of the problem)

Also you know AC+BC=AB (See the figure you drew!)

How far behind is second one is actually asking what's the distance between the two which is BC (See the figure !!!!)

AB=62

AC( you found above)

Find BC using AC+BC=AB

You will get BC in cms ( If you divide it by 100 that would be in m)

Hope this helps, I wish I could draw the figure for you.