• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

L

librement

Messages
5
Reaction score
2
Points
13
can someone help me out with this question, part (ii) only? Thanks!
 

Attachments

  • Screen Shot 2015-02-17 at 7.13.42 pm.png
    Screen Shot 2015-02-17 at 7.13.42 pm.png
    59.7 KB · Views: 6
I

iqbal

Messages
18
Reaction score
6
Points
13
sagar65265 said:
4) i)

Since we know that the speed of each particle is the same as that of the other when Q has reached point B and P has reached point A, our plan of action can be equating the speeds of the particles at those points, and using that equation to find the initial velocity of Q.

Firstly, we know that Q starts out with an unknown initial velocity and accelerates with a non-uniform acceleration magnitude toward point B, while P starts out with a known initial velocity and accelerates with uniform acceleration magnitude toward point A. We also know that the time taken for Q to reach B is the same as the time taken for P to reach A.

Armed with these facts, let's begin.

Since P moves with a constant acceleration, we can say that the velocity of P at point A will be given by

v = u + at

and putting the values we know (a = 0.1 ms^-2 , u = 1.3 ms^-1 and t = 20 seconds), we get

v = 1.3 + 0.1 * 20 = 3.3 ms^-1

As for Q, we can write the acceleration equation as

a = dv/dt = (0.016t) ms^-2

which has the variable "t" on the right hand side, and a derivative with respect to this very variable on the left hand side - a differential equation. If we multiply both sides by "dt", we get

dv = (0.016t) * dt

and this we can integrate - on the right hand side, we integrate with respect to time, on the left hand side, with respect to velocity, giving us

v = 0.016 * (t2/2) + C

Where C is a constant representing initial conditions (if you put t = 0, you get v = C). Substituting t = 20 seconds in this equation, we get

v = 0.016 * (400/2) + C

We know that the velocity of particle Q at point B is the same as the velocity of particle P at point A, and since we calculated the velocity of particle P at point A to be 3.3 ms^-1, we can insert this value of v into the left hand side, giving us

3.3 = 3.2 + C
Therefore, C = 3.3-3.2 0.1 ms^-1. Therefore, our final equation is

v(t) = 0.008t2 + 0.1

and if we put t = 0 seconds , we get v(0) = 0.1 ms^-1, our answer to part (i).

4) ii)

Here, since P travels from O to A and Q travels from O to B, if we find the distance P travels in 20 seconds (when it reaches A) and add it to the distance Q travels in 20 seconds (when it reaches B) we have found the total distance from A to B, which is the length AB we are looking for.

Since P travels with a constant acceleration, we can find the distance it travels with the equation

s = ut + (1/2)at2

which gives us

s = (1.3)(20) + 0.5*0.1*400 = 46 meters

Which is the length OA. Now for length OB, we need to find the distance traveled by the particle Q, which we can get from the equation v(t) = 0.008t2 + 0.1, by writing it as

v(t) = dx/dt = 0.008t2 + 0.1

and if we multiply both sides by dt, we get

dx = (0.008t2 + 0.1) * dt

And on integrating, we get

x = 0.008 * (t3/3) + 0.1t + C

and if we put t=0 seconds and x = 0 meters (at t = 0), we see that C = 0. Therefore,

x(t) = 0.008 * (t3/3) + 0.1t

and if we put t = 20 seconds into this equation (which is when Q reaches B) we get

x(20) =23.33 meters.

So adding these two values of x, we get 23.33 + 46 = our answer = 69.33 m.

Hope this helped!

Good Luck for all your exams!
Click to expand...
Thanks
 
I

iYuuki

Messages
10
Reaction score
3
Points
13
Hey guys, could someone help me out with question 6 (ii)?? Really appreciate it! >.<
 

Attachments

  • P3.jpg
    P3.jpg
    32 KB · Views: 5
AnujaK

AnujaK

Messages
117
Reaction score
215
Points
53
SitiPutri said:
Hi. Would've been better if you could compile all the images in one pdf file for each exam. It'd be much easier, instead of downloading the images one by one. Thanks for making the site, btw. :)
Click to expand...
Thanks for the tip! I'll try that :)
 
Haya Ahmed

Haya Ahmed

Messages
354
Reaction score
529
Points
103
Zain Salman Dar said:
Draw a circle of radius 2 cm at (0,2)
Draw an angle of 45 degrees (-2,0)
Calculate distance between (-2,0) and the circle+diameter
Click to expand...
how and why did you calculate the distance for modZ ... isn't it supposed to be from the (0,0) to the farthest point on the shaded region ?!
 
Haider Ejaz

Haider Ejaz

Messages
28
Reaction score
2
Points
13
MATH HELP NEEDED...PLEASE HELP

How to solve this inequality question?
(3x+1)(x+1)>0

Truly its an easy question but i am confused that if we take (3x+1)>0 and (x+1)>0 then it shows answer wrong.... why is this so??????? please help
 
F

Fatma Gahman

Messages
43
Reaction score
13
Points
18
A student is chosen at random. The probability that the student estimates that the mass is greater than M grams is 0.3. Find the value of M.?? HELP PLZZ
 
You must log in or register to reply here.
Top