# Physics problem ...

• Nov 3rd 2010, 02:10 AM
r-soy
Physics problem ...
• Nov 3rd 2010, 10:48 AM
Unknown008
1. Remember that : $Volume = length \times width \times height$ and the unit is $m^3$ and not m.

2. Right.

3. a) Right
b) ... what can I say here... you used the rounded value you got previously, and this is often a very risky thing to do... The safest way is to use what you were given and what you are sure. For instance, the internal angle in a triangle is 180 degrees.

The ladder, the ground and the wall make up the three sides of a right angle triangle (sketches always help). Hence, the angle the wall makes with the ladder becomes (180 - 90 - 65) = 25 degrees exactly, and not 25.8 degrees.

4.

a) No, that's not it. The displacement is given by the area under the graph. Can you try this again?

b) Right.

c) Um... the gradient at a point gives the acceleration, right? What is the gradient at t = 4s?

5. Right... the question is asking you for the distance required for stopping completely.

I'll advise you to find the acceleration if you cannot 'play' with variables without making mistakes.

Use the equation: $v^2 = u^2 + 2as$

For the car to stop, we get v = 0. We know that u = 15.6, s = 12.2.

You can find a.

Then, use the same equation to get the distance for the higher speed.

This time, you know a, you are given vo and you have to find s, knowing tha v = 0

Post what you get! (Smile)
• Nov 4th 2010, 11:14 PM
r-soy

How i find s(t)

and to find Did i use equation: http://www.mathhelpforum.com/math-he...d484645924.png

help me
• Nov 4th 2010, 11:42 PM
Unknown008
Uh... what do you mean?

From $v^2 = u^2 + 2as$

We get:

$0 = (15.6)^2 + 2a(12.2)$

$a = \dfrac{-243.36}{2(12.2)} = - 9.97 m/s^2$

Using the same equation for the other case,

$0 = (31.2)^2 + 2(-9.97)s$

$s = ?$

I get 48.8 m
• Nov 5th 2010, 02:31 AM
r-soy
Hi but from where you got the answer of time = 12.2

and this is all answer ?

?
• Nov 5th 2010, 06:43 AM
Unknown008
s here is not the time but the displacement. This is the equivalent of 40 ft that you were given and that you yourself worked out.
• Nov 5th 2010, 11:16 AM
r-soy
Ok about 5 now we say 4 X 40 = 160 ft

part a did the answer is 65
• Nov 5th 2010, 11:42 AM
Unknown008
Quote:

Originally Posted by r-soy
Ok about 5 now we say 4 X 40 = 160 ft

part a did the answer is 65

Each square has an area of 2*5 = 10 m

Counting everything, there are 7.5 squares, hence 75 m.

(6 full squares, 2 squares divided into two makes one whole square and half a square)

5. I'm not sure what you are telling me. The final answer should be 48.8 m
• Nov 5th 2010, 12:48 PM
r-soy
In queation 5

I got s = 48.8

then now how i get for the minimum stopping distance ?
• Nov 5th 2010, 12:49 PM
Unknown008
Um... 48.8 m is the minimum stopping distance... you can convert it into feet now.
• Nov 5th 2010, 01:27 PM
r-soy
yes, now ok

I got 160
• Nov 8th 2010, 08:59 AM
r-soy
Sand dunes on a desert island move as sand is swept up the windward side to settl
1 ) Sand dunes on a desert island move as sand is swept up the windward side to settle in the leeward side. Such “walking” dunes have been known to travel 20 feet in a year and can travel as much as 100 feet per year in particularly windy times. Calculate the average speed in each case in m/s.
(b) Fingernails grow at the rate of drifting continents, about 10 mm/yr. Approximately how long did it take for North America to separate from Europe, a distance of about 3 000 mi?
a )
20ft = 32180m
average speed = 32180/100 = 321.8
b)
t = 3000/10 = 300

---------------------------------------

2 ) A bristlecone pine tree has been know to take 4000 years to grow to a height of 20 ft.
a)Find the avearge speed of growth in m/s
b) In contrast the faster growing palnt is the gaint kelp which can grow at rate of 2 feet in one day
a )
4000 years conver to secound
=345.6X10^-6
now 20 ft to m = 6.1
avearge speed = 6.1/345.6X10^-6 = 1.7650
b) I don't understand this ?

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3 ) Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 70 mi/h. (a) Assuming that they start at the same point, how much sooner does the faster car arrive at a destination 10 mi away? (b) How far must the faster car travel before it has a 15-min lead on the slower car?

give me the idea in this queation .
• Nov 8th 2010, 09:34 AM
Unknown008
1. a) I trust you for the conversions.

It covers 20 ft in a year.

A year contains (356.25 x 24 x 3600) seconds [a year -> 365.25 days; 1 day -> 24 hours; 1 hour -> 3600 seconds]. Divide the distance the dune travels by this.

Now, it covers 100 ft in a year. The average speed then becomes the distance in metres of 100 ft divided by that number of seconds.

b) The units are not consistent! Convert the 3000 mi into mm first, then do this calculation.

2. a) Uh... 1 year alone contain more than one second, how can 4000 years contain less than 1 second? The number of seconds is wrong.

b) I think you need to find the speed of growth of this giant kelp too...

3.

a) Find the time it takes for the faster car to cover 10 mi.

Find the time it takes for the slower car to cover 10 mi.

Find the difference in time.

b) Set up two equations. One for the slower car and one for the faster car, in terms of their speed (v1 and v2), distance travelled (d) and time travelled (h), the time travelled being the subject of the formulae.

For the faster car, it travels a distance d in time h hours
For the slower car, it travels the same distance d but in time (h + 0.25) hours.

You get two equations with two unknowns. Solve for d, the distance covered by the faster car.