problem on time and speed

Topic :- Time and Speed


Time problems and speed problems are there in both math and Physics .Here is a simple math problem which help you to understand to how to find time when speed value is given . This problem solved
by using distance formula ,So first we find the distance based on which we get the time .

Question:-
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
Answer:-
Let the time taken by the second cyclist to catch up the first be 'x' hours.

Then the time taken by the first to reach than the point wil be (x+3) hours.

Speed of the second cyclist=10 miles/hour.

So the distance covered =speed*time
= 10 * x = 10x miles

Speed of the first cyclist = 6 miles/hour

So distance covered = 6(x+3)
=(6x+18) miles

Now distnce covered by both are same .

So

10x = 6x+18
-6x   -6x
------------------
4x  =  18

4x     18
---- = ----
4      4

   18     9        1
x= ---- = ---- = 4----
   4     2         2

So the time taken by the second
cyclist = 4 hours 30 minutes

I hope you liked it ,For math help ,you can reply me

Question on Circles, Finding unknown values of attributes of the circle

Circle is a geometrical figure with basic attributes diameter, radius, circumference, arc, chord etc. Below is a Geometry construction for the circle and to find unknown attributes value.

Topic : Circle and its attributes

Tangent is a straight line which just touches the circumference of the circle at a single point.

Question : Given VQ is tangent to the circle O at Q, QS is a diameter of the circle O, arc PQ = 115⁰, angle RPS = 36⁰

Find a) angle R ; b) angle S ; c) arc SR ; d) arc QR ; e) angle QPR ; f) angle QPS ; g) angle QTP











Solution :











Given : mPQ = 115⁰
‹RPS = 36⁰

a) ‹R
Since QS is the diameter,
So ‹QRS = 90⁰
(as angle in a semi circle is a right angle)

b) ‹S
As mPQ = 115⁰
So ‹PSQ = 1/2 * (115⁰) = 57.5⁰
(angle subtented by an arc at any point on the circle is half of the angle subtented by it at the center)

c) mSR
given ‹SPR = 36⁰
So mSR = 2(36) = 72⁰

d) mQR
now mQS = 180⁰
mQR = mQS - mSR
mQR = 180⁰ - 72⁰
mQR = 108⁰

e) ‹QPR
mQR = 108⁰
So ‹QPR = 1/2*(108) = 54⁰

f) ‹QPS
‹QPS = 90⁰(as QS is the diameter)

g) ‹QTP
mQP = 115⁰
so ‹QSP = 1/2 *(115⁰) = 57.5⁰
now in triangle PTS
‹QTP = 180⁰ - (‹TPS + ‹PST)
‹QTP = 180⁰ - (36⁰ + 57.5⁰)
‹QTP = 180⁰ - 93.5⁰
‹QTP = 86.5⁰

Well these are the bits of circle, to get more examples and get practiced, contact geometry help.

A Simple Word Problem on Finding an Integral of an Expression/Temperature vs Time problem

Here is a Multi choice word problem on Temperature and Time to Evaluate the temperature by Integration. Also converting the value into degrees. You can find similar set of questions at TutorVista Blogs.

Topic : Integration.

Calculation in the solution will help you to identify correct choice and get reasons for, why remaining choices are incorrect.

Question : An observer measures the outside temperature every hour from noon until midnight, recording the temperatures in the following table
Time Temp
N -----63
1 ----- 65
2 ----- 66
3 ----- 68
4 ----- 70
5 ----- 69
6 ----- 68
7 ----- 68
8 ----- 65
9 ----- 64
10 ---- 62
11 ---- 58
M ----- 55

Then the average temperature for the 12-hour period is

a. 71
b. 76
c. 65
d. 66

Solution :
Choice c is correct.

We are looking for the average value of a continuous function (temp.) for which we know values at discrete times that are one unit apart.
We need to find
Average (f) = 1/(b-a)_a∫^bf(x)dx
Without having a formula for f(x)
The integral, however can be approximated by the Trapezoidal Rule, taking h=1

T = h/2(y₀ + 2y₁ + 2y₂ + ..... + 2y₁₁ + 2y₁₂)

= 1/2 (63 + 2*65 + 2*66 + ...... + 2*58 + 2*55)

= 782

average (f)approximately 1/(b-a)T = 1/12*782 = 65.17
Therefore, the average temperature =65 degrees

Choice a is incorrect because of the error in considering the interval between noon to midnight as 13 hours instead of 12 hours.

Choice b is incorrect because of the error in considering the interval between noon to midnight as 11 hours instead of 12 hours.

Choice d is incorrect because of error in calculating the trapezoidal approximation T as 792 instead of 782.

If you have any queries related to this problem or Integration, please do write to us and you will surely get help from calculus help

Question to Find Percentage of Discount on a Calculator Price

Topic : Sales and Price

Question : Peter bought a calcuator on sale for $40 the calculators retail price is $60 what is the percent discount?

Solution :
Discount in Price = 60 - 40 = $20
So discount percentage = discount price /Retail Price * 100
= 20/60 * 100
= 100/3
= 33.33%

Problem on Trigonometry Equation and Proof for the Identities

Topic : Trigonometry Identites

Simplify.

Problem on Trigonometric Equations

Topic : Trigonometric Equation

Problem : Solve 3Sin θ Cos θ – 2 Cos θ = 0 and 0 ≤ θ ≤ 2π

Solution :

3Sin θ Cos θ – 2 Cos θ = 0
Taking Cos θ as common
Cos θ (3Sin θ - 2) = 0
Cos θ = 0 or 3Sin θ – 2 = 0
3 Sin θ – 2 = 0
3Sin θ = 2
Sin θ = 2/3 (2/3 = 0.66 = 41.8º ~ 42º)
Sin θ = Sin 42º
θ = n π + (-1)ⁿ 42º
when n = 0 , θ = 42º
when n = 1 , θ = π - 42º = 138º

Let’s solve Cos θ = 0
Cos θ = Cos π/2
The formula for Cosine is :
If Cos x = Cos y
Then x = 2nπ ± y

So here it will be θ = 2nπ ± π/2
When n = 0, θ = 2(0)π ± π/2 = ± π/2
But only + π/2 lies in 0 ≤ θ ≤ 2π
So x = π/2

When n = 1, θ = 2(1)π ± π/2 = 2π ± π/2 = 2π ± π/2 = (4π ± π)/2 = 3π/2, 5π/2
But x = 3π/2

Now the further values of n will give angles greater than 360º

So θ will be 42º, 90º, 138º, 270º

Problem on Simple Linear Equation

Topic : Linear Equation

Problem : Solve y = 3y - 12

Solution :